Show that $(\sqrt{y^2-x}-x)(\sqrt{x^2+y}-y)=y \iff x+y=0$ 
Let $x,y$ be real numbers such that
  $$\left(\sqrt{y^{2} - x\,\,}\, - x\right)\left(\sqrt{x^{2} + y\,\,}\, - y\right)=y$$
  Show that $x+y=0$.

My try:
Let
$$\sqrt{y^2-x}-x=a,\sqrt{x^2+y}-y=b\Longrightarrow ab=y$$
and then
$$\begin{cases}
y^2=a^2+(2a+1)x+x^2\cdots\cdots (1)\\
x^2=b^2+(2b-1)y+y^2\cdots\cdots
\end{cases}$$
$(1)+(2)$
then
$$x=-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}\cdots\cdots (3)$$
so
$$x+y=ab-\dfrac{a^2+b^2+(2b-1)ab}{2a+1}=\dfrac{(a-b)(2ab-a+b)}{2a+1}$$
we take $(3)$ in $(2)$,we have
$$b^2+(2b-1)y+y^2-x^2=\dfrac{(2ab-a+b)(2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b)}{(2a+1)^2}=0$$
so
$$(2ab-a+b)=0$$
or
$$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$
if
$$2ab-a+b=0\Longrightarrow x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0$$
and if
$$2a^3b+a^3+3a^2b-2ab^3+ab^2+4ab-b^3+b=0$$
I don't prove
$$x+y=\dfrac{(a-b)(2ab-a+b)}{2a+1}=0?$$
 A: Assume that $x<0$ and $y>0$, your statement can be written (even if we change $x$ to $-x$),

If $x,y$ are strictly positive such that $(\sqrt{y^2+x}+x)(\sqrt{x^2+y}-y)=y$ then $x=y$.

This equality becomes, $$\overbrace{\dfrac{x+\sqrt{x+y^2}}{y+\sqrt{y+x^2}}}^{A}=\overbrace{\dfrac{y}{x^2+y-y^2}}^{B}.$$ Let us start with $x>y$, clearly $B<1$. Moreover $x\mapsto(x-\sqrt{y+x^2})$ is increasing and $x\mapsto \sqrt{y+x^2}$ strictly increasing, we conclude that $x\mapsto(x+\sqrt{x+y^2})-(y+\sqrt{y+x^2})$ is strictly increasing and thus $A>0$ when $x>y$. We deduces that $A>1$ and therefore $A$ cannot be equal to $B$.
The case $x<y$ can be treated in the same way.
A: Multiplication: 
$y=x y-x \sqrt{x^2+y}-y \sqrt{-x+y^2}+\sqrt{x^2+y} \sqrt{-x+y^2} \Rightarrow$
$x \sqrt{x^2+y}+y \sqrt{-x+y^2}=\sqrt{x^2+y} \sqrt{-x+y^2}+xy-y$  
squaring both sides:
$2 x y \sqrt{x^2+y} \sqrt{y^2-x}+x^4+x^2 y-x y^2+y^4= 
-x^3-x y+y^2-2 x y^2+2 x^2 y^2+y^3+(2 x y-2 y  )\sqrt{x^2+y} \sqrt{-x+y^2} \Rightarrow $
simplifying:
$x^4+x^3-2 x^2 y^2+x^2 y+x y^2+x y+y^4-y^3-y^2 = -2 y \sqrt{x^2+y} \sqrt{y^2-x}$
Squaring again:
$y^8 - 2 y^7 - 4 x^2 y^6 + 2 x y^6 - y^6 + 6 x^2 y^5 + 2 y^5 + 6 x^4 y^4 - 2 x^3 y^4 + 3 x^2 y^4 - 4 x y^4 + y^4 - 6 x^4 y^3 - 4 x^3 y^3 - 2 x y^3 - 4 x^6 y^2 - 2 x^5 y^2 + x^4 y^2 + x^2 y^2 + 2 x^6 y + 4 x^5 y + 2 x^4 y + x^8 + 2 x^7 + x^6=
-4 x^3 y^2+4 x^2 y^4-4 x y^3+4 y^5$
The only thing we have to check is how it factors as $(y+x) (p(x,y))$. Which gives us:
$(x+y)^2(x^6-2 x^5 y+2 x^5-x^4 y^2-2 x^4 y+x^4+4 x^3 y^3+2 x^3 y-x^2 y^4-4 x^2 y^3-4 x^2 y^2+2 x^2 y-2 x y^5+6 x y^4+2 x y^3+y^6-2 y^5-y^4-2 y^3+y^2)=0$
A: In the following proof we divide the $(x,y)$ plane into regions (see diagram) and show that each region can contain no solutions except on the line $x + y = 0$.

Let
$$
F(x,y) = U(x,y)V(x,y) – y
$$
where
$$
U(x,y) = \sqrt{y^2 – x} \,\, – x
\\V(x,y) = \sqrt{x^2 + y} \,\, – y
$$
Then solutions satisfy
$$
F(x,y) = 0
$$
Substituting $y=-x$ shows that $x+y=0$ is a solution for all x.
U is nonreal where $x>y^2$ (regions A and C in the diagram, bounded by red lines) and V is nonreal in the region $y<-x^2$ (regions B and C, also bounded by red lines). In these regions F is nonreal except possibly in region C, where U and V are both unreal. But there the condition for F to be real reduces to $x+y=0$, a subset of the known solution.
The following statements and deductions relate to the other regions of the $(x,y)$ plane, where $U$ and $V$ are real.
$U<0 \Leftrightarrow x>\tfrac{1}{2}(-1 + \sqrt{1 + 4y^2})$ (regions I, J).
$V<0 \Leftrightarrow y>\tfrac{1}{2}(1 + \sqrt{1 + 4x^2})$ (all regions except F, G).
$U_{x} < 0$ (regions D-K)
$V_{x} < 0 \Leftrightarrow x < 0$ (regions D, E, F)
$U_{xx} < 0$ (regions D-K)
$V_{xx} < 0 \Leftrightarrow y < 0$ (regions D, J, K)
where a subscript x denotes partial differentiation with respect to x.
On the diagram the lines on which $U=0$ and $V=0$ are coloured green and blue, respectively. It is easily shown that $F$ is nonzero on all the coloured lines (with sign as indicated) except at $(0,0)$ and $(1,-1)$. These lines delimit, but are excluded from, the regions A-K.
From the results above we can make the following deductions.
In region D:
$$
U>0, V>0, U_{x}<0, V_{x}<0
\\F_{x} = UV_{x} + VU_{x} < 0
$$
This region is bounded on the right by the line $y<-x^2$, on which $F>0$. So $F>0$ throughout region D and it can contain no solutions.
In region E:
$$
U>0, V>0, U_{x}<0, V_{x}<0
\\F_{x} < 0
$$
so here there can be no solutions other than those known to exist on the line $y=-x$.
In region F:
$$
U>0, V<0, U_{x}<0, V_{x}<0, U_{xx}<0, V_{xx}>0
\\F_{xx} = UV_{xx} + VU_{xx} + 2U_{x}V_{x} > 0
$$
This region is bounded on the left by the line $V=0$ and on the right by the line $x = 0$, and on both these lines $F<0$. So the positive second derivative $F_{xx}$ means there can be no solutions $F=0$ in this region.
In a similar way, solutions can be ruled out for the following regions:
In region G, bounded on right by the line $V=0$ on which $F<0$:
$$
U>0, V<0, U_{x}<0, V_{x}>0
\\F_{x}>0
$$
In region I, bounded on left by the line $U=0$ on which $F<0$:
$$
U<0, V>0, U_{x}<0, V_{x}>0
\\F_{x} < 0
$$
In region J, containing a segment of the known solution line $x+y=0$ on which $F=0$:
$$
U<0, V>0, U_{x}<0, V_{x}>0
\\F_{x} < 0
$$
In region K, bounded on the left and right by lines on which $F>0$:
$$
U>0, V>0, U_{x}<0, V_{x}>0, U_{xx}<0, V_{xx}<0
\\F_{xx} = UV_{xx} + VU_{xx} + 2U_{x}V_{x} < 0
$$
Finally, in region H:
$$
U>0, V>0, U_{x}<0, V_{x}>0
$$
and we note that $U_{x}<0$ in region G also, so for a given $y$,
$U<U_{max}$, where $U_{max} = U(0,y) = y$
For the same value of y, $V<V_{max}$, where $V_{max} = V(X,y)$, and X is the value of x on the right-hand boundary of the region. On this boundary, $y=\sqrt{X^2+X}$, so
$$
V_{max} = V(X,y) = \sqrt{X^2+y} \, – y
< \sqrt{X^2+X+y} \, - y = \sqrt{y^2 + y} \,\, – y < \tfrac{1}{2}.
$$
Therefore
$$
F = UV – y < U_{max} V_{max} – y  < y \tfrac{1}{2} – y = \, –\tfrac{1}{2} y < 0
$$
which completes the proof that there are no solutions other than $x+y = 0$.
A: Assuming continuity in the area interval $ (0 < x < 1 )$ and $ (0 > y > -1  ) $ would create problems as $x$ and $y$ are not always real in these areas.
Nay, union of inside parabola areas of $ y_1 = - x^2 $  and $  y_2 = \sqrt{x} $ would violate $ x + y = 0, $ which is only the common chord of intersection of  $ y_1,y_2$. So the shown line joining $(0,0)$ to $(1,-1)$ does not exist as real.
E.g., $(\frac12,  -\frac12)$ does not lie on the
 common line.

A: This is not a solution, but brute force can be used to remove the radicals. Let $A=y^2-x$ and $B=x^2+y$. We have
$$\sqrt{AB}-y\sqrt{A}-x\sqrt{B}+xy=y$$
Isolating $\sqrt{AB}$ and squaring both sides:
$$\sqrt{AB}=y\sqrt{A}+x\sqrt{B}+y(1-x)\quad(1)$$
$$AB=y^2A+x^2B+y^2(1-x)^2+2xy\sqrt{AB}+2y^2(1-x)\sqrt{A}+2xy(1-x)\sqrt{B}$$
(1) allows us to remove $\sqrt{AB}$. We do this and also recall what $A$ and $B$ equal.
$$(y^2-x)(x^2+y)=y^2(y^2-x)+x^2(x^2+y)+y^2(1-x)^2+2xy\left(y\sqrt{A}+x\sqrt{B}+y(1-x)\right)+2y^2(1-x)\sqrt{A}+2xy(1-x)\sqrt{B}$$
Group $\sqrt{A}$ and $\sqrt{B}$ terms, then rearrange a bit:
$$(y^2-x)(x^2+y)=y^2(y^2-x)+x^2(x^2+y)+y^2(1-x)^2+2xy^2(1-x)+2y^2\sqrt{A}+2xy\sqrt{B}$$
$$x^2y^2+y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2xy^2+x^2y^2+2xy^2-2x^2y^2+2y\left(y\sqrt{A}+x\sqrt{B}\right)$$
$$y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2x^2y^2+2y\left(y\sqrt{A}+x\sqrt{B}\right)$$
(1) allows us to sub out the quantity in parentheses:
$$y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2x^2y^2+2y\left(y(x-1)+\sqrt{AB}\right)$$
$$y^3-x^3-xy=y^4-xy^2+x^4+x^2y+y^2-2x^2y^2+2y^2(x-1)+2y\sqrt{AB}$$
$$y^3-x^3-xy=y^4+xy^2+x^4+x^2y-y^2-2x^2y^2+2y\sqrt{AB}$$
$$y^3-x^3-xy-y^4-xy^2-x^4-x^2y+y^2+2x^2y^2=2y\sqrt{AB}$$
Squaring both sides, we've reached a goal of no longer having radicals.
$$(y^3-x^3-xy-y^4-xy^2-x^4-x^2y+y^2+2x^2y^2)^2=4y^2(y^2-x)(x^2+y)$$
I had a CAS expand this, move it all to one side, and then, as expected, $(x+y)$ factors out of it (twice).
$$(x+y)^2 p(x,y)=0$$
where $$p(x,y)=x^6-2 x^5 y+2 x^5-x^4 y^2-2 x^4 y+x^4+4 x^3 y^3+2 x^3 y-x^2 y^4-4 x^2 y^3-4 x^2 y^2+2 x^2 y-2 x y^5+6 x y^4+2 x y^3+y^6-2 y^5-y^4-2 y^3+y^2$$ 
is a monster. It would be sufficient to show that $p(x,y)$ is never $0$ in the region of the plane where both $\sqrt{A}$ and $\sqrt{B}$ are defined aside from points along $x+y=0$ (like $(0,0)$). This is a pretty messy polynomial, but at least it's a polynomial.

EDIT: This approach seems to be useless; a CAS plot of the zero set of $p$ has several components, all of which are in the region where $\sqrt{A}$ and $\sqrt{B}$ are defined. They must be extraneous solutions from the squaring that was done twice.
A: A more generalised approach over my earlier post. This is not intended to be an exhaustive proof but an experimental one. Constructive comments are most welcome. 
Let 
$$\sqrt{y^2-x}-x=Ay^n\qquad \cdots (1)\\
\sqrt{x^2+y}-y=\frac {y^{1-n}}A \qquad \cdots (2)\\$$
such that the original equation $$\left(\sqrt{y^2-x}-x\right)\left(\sqrt{x^2+y}-y\right)=y$$
is satisfied as required.
From $(1)$,
$$\begin{align}
\sqrt{y^2-x}&=x+Ay^n\\
y^2-x&=x^2++2Axy^n+A^2y^{2n}\\
y^{2n}A^2+2xy^nA+(x^2-y^2+x)&=0\\
A^2+\frac {2x}{y^n}A+\frac{(x^2-y^2+x)}{y^{2n}}&=0\qquad \qquad \qquad \qquad \cdots (3)
\end{align}$$
From $(2)$,
$$\begin{align}
\sqrt{x^2+y}&=y+\frac {y^{1-n}}A\\
A\sqrt{x^2+y}&=Ay+y^{1-n}\\
A^2(x^2+y)&=A^2y^2+2Ay^{2-n}+y^{2(1-n)}\\
(x^2-y^2+y)A^2-2y^{2-n}A-y^{2(1-n)}&=0\\
A^2-\frac{2y^{2-n}}{x^2-y^2+y}A-\frac{y^{2(1-n)}}{x^2-y^2+y}&=0\qquad \qquad \qquad \cdots (4)
\end{align}$$
Equating coefficients of $A^1$:
$$\begin{align}\frac{2x}{y^n}&=-\frac{2y^{2-n}}{x^2-y^2+y}\\
y^2&=-x(x^2-y^2+y)\qquad \qquad \qquad \qquad \qquad \qquad \cdots (5)\end{align}$$
Equating coefficients of $A^0$:
$$\begin{align}\frac{x^2-y^2+x}{y^{2n}}&=-\frac{y^{2(1-n)}}{x^2-y^2+y}\\
y^2&=-(x^2-y^2+x)(x^2-y^2+y)\qquad \cdots (6)\end{align}$$
(5)=(6):
$$\begin{align}x(x^2-y^2+y)&=(x^2-y^2+x)(x^2-y^2+y)\\
(x^2-y^2)(x^2+y^2-y)&=0\\
(x-y)(x+y)(x^2-y^2+y)&=0\\
\Rightarrow x-y=&0, x+y=0, x^2-y^2+y=0\end{align}$$
Checking by substitution into the original equation shows that only 
$$x+y=0$$
is valid.
This graph created on desmos.com might help illustrate the approach:
https://www.desmos.com/calculator/qrlbgbalix
A: We need to prove that $x=y$, where
$$\left(\sqrt{y^{2}+x}+x\right)\left(\sqrt{x^{2} + y}- y\right)=y$$ or
$$\left(\sqrt{y^{2}+x}+x\right)\left(\sqrt{x^{2} + y}- y\right)=\left(\sqrt{x^{2}+y}+x\right)\left(\sqrt{x^{2} + y}- x\right)$$ or
$$\left(\sqrt{y^{2}+x}+x\right)\left(\sqrt{x^{2} + y}- y\right)-\left(\sqrt{x^{2}+y}+x\right)\left(\sqrt{x^{2} + y}- y\right)+$$
$$+\left(\sqrt{x^{2}+y}+x\right)\left(\sqrt{x^{2} + y}- y\right)-\left(\sqrt{x^{2}+y}+x\right)\left(\sqrt{x^{2} + y}- x\right)=0$$ or
$$\left(\sqrt{y^{2}+x}-\sqrt{x^2+y}\right)\left(\sqrt{x^{2} + y}- y\right)+\left(\sqrt{x^{2}+y}+x\right)(x-y)=0,$$ which gives $x=y$ or
$$\frac{(1-x-y)\left(\sqrt{x^{2} + y}- y\right)}{\sqrt{y^{2}+x}+\sqrt{x^2+y}}+\sqrt{x^{2}+y}+x=0,$$ which is
$$\sqrt{x^2+y}-y\sqrt{x^2+y}+\sqrt{(x^2+y)(y^2+x)}+x\sqrt{y^2+x}+x^2+xy+y^2=0.$$
Now, we'll consider four cases.


*

*$x\geq0$, $y\geq 0$.


Since $$-y\sqrt{x^2+y}+\sqrt{(x^2+y)(y^2+x)}=\sqrt{x^2+y}\left(\sqrt{y^2+x}-y\right)\geq0,$$ we obtain $x=y=0.$


*$x\geq0,$ $y\leq0.$
It's obvious that this case gives $x=y=0$ again.


*$x\leq0$, $y\geq0.$
Since, $$\sqrt{(x^2+y)(y^2+x)}+x\sqrt{y^2+x}=\sqrt{y^2+x}\left(\sqrt{x^2+y}+x\right)\geq0,$$ it's enough to prove that
$$x^2+xy+y^2\geq(y-1)\sqrt{x^2+y},$$ which is obvious for $y\leq1.$
But for $y\geq1$ by AM-GM we obtain:
$$(y-1)\sqrt{x^2+y}\leq\frac{1}{2}((y-1)^2+x^2+y)$$ and it's enough to prove that
$$x^2+xy+y^2\geq\frac{1}{2}((y-1)^2+x^2+y)$$ or
$$\require{cancel} \cancel{(x+y)^2+y^2+y-1\geq0.}\\
(x+y)^2+y-1\geq0.$$
We see that for $y\geq1$ the equality does not occur and in the case $y<1$ the equality occurs for
$$x^2+xy+y^2=(y-1)\sqrt{x^2+y}=0,$$ which gives $x=y=0$ again. 


*$x\leq0$ and $y\leq0.$
In this case it's enough to prove that
$$xy+x\sqrt{y^2+x}\geq0$$ or
$$x\left(y+\sqrt{y^2+x}\right)\geq0,$$ which is obvious.
The equality occurs for $x^2+y^2=0$ and we got $x=y=0$ again.
Done!
