If $x\sqrt{1+y}+y\sqrt{1+x}=0$ find $y'$ 
Find $\frac{dy}{dx}$ if $x\sqrt{1+y}+y\sqrt{1+x}=0$ for $-1\leq x\leq 1$

My Attempt
$$
x\sqrt{1+y}=-y\sqrt{1+x}\implies x^2(1+y)=y^2(1+x)\implies x^2+x^2y=y^2+xy^2\\
2x+2xy+x^2\frac{dy}{dx}=2y\frac{dy}{dx}+y^2+2xy\frac{dy}{dx}\\
\frac{dy}{dx}\Big[ x^2-2y-2xy \Big]=y^2-2x-2xy\\
\frac{dy}{dx}=\frac{y^2-2x-2xy}{x^2-2y-2xy}
$$
How do I proceed further and find the derivative ?
 A: Hint:
Squaring both sides of $$x\sqrt{1+y}=-y\sqrt{1+x}$$ 
$$x^2(1+y)=y^2(1+x)$$
$$\iff0=(x-y)(x+y+xy)$$
But $x\sqrt{1+y}=-y\sqrt{1+x}\implies x,y$ are of opposite sign, hence $x\ne y$ unless $x=y=0$
Otherwise, $$y(1+x)=-x\implies y=\dfrac{1-(x+1)}{1+x}=\dfrac1{1+x}-1$$
A: $x^2+x^2y = y^2 + xy^2 $
$x^2+x^2y -y^2-xy^2 = 0 $ 
$(x-y)(x+y) + xy(x-y) = 0 $ 
$(x-y)(x+xy+y) = 0 $ 
$(x-y) = 0 , or, x + xy + y = 0 $
$x \neq y$ since $x\sqrt{1+y} = -y\sqrt{1+x}$ unless (0,0)
$x + xy + y = 0 $
$x + y(x + 1) = 0 $
$y = \frac{-x}{x+1} = -1 + \frac{1}{x+1}$
$\frac{dy}{dx} = -(x+1)^2$
A: $$
x\sqrt{1+y}+y\sqrt{1+x}=0\implies x\sqrt{1+y}=-y\sqrt{1+x}\\
\implies x^2(1+y)=y^2(1+x)\implies x^2+x^2y=y^2+y^2x\\
\implies \boxed{(1+x)\color{red}{y^2}-x^2\color{red}{y}-x^2=0}
$$
$B^2-4AC=x^4+4x^2(1+x)=x^4+4x^2+4x^3=x^2(x^2+4x+4)=x^2(x+2)^2$
So,
$$
\begin{align}
y&=\frac{x^2\pm\sqrt{x^2(x+2)^2}}{2(x+1)}=\frac{x^2\pm|{x(x+2)}|}{2(x+1)}\\
&=\frac{x^2+x(x+2)}{2(x+1)}\text{ or }\frac{x^2-x(x+2)}{2(x+1)}\\
&=\frac{x^2+x^2+2x}{2(x+1)}\text{ or }\frac{x^2-x^2-2x}{2(x+1)}\\
&=\frac{2x(x+1)}{2(x+1)}\text{ or }\frac{-2x}{2(x+1)}\\
y&=x \text{ or } \frac{-x}{x+1}
\end{align}
$$
As $y\neq x$ we have the derivative,
$$\color{blue}{
y'=\frac{-1}{(1+x)^2} }
$$
Note: $x+\sqrt{1+y}=−y\sqrt{1+x}\implies(x−y)(x+y+xy)=0\implies x−y=0$ or $(x+y+xy)=0$. Note that $x−y=0\implies x=y$ is equivalent to the original function only when $x=0$, and $x\sqrt{1+y}=−y\sqrt{1+x}\Leftrightarrow(x+y+xy)=0$ for all $x∈D_\text{original function}$
