$x^2+y^2+9=3(x+y)+xy$ Find all pairs of real $x,y$ that meet this equation $\frac{(x-y)^2}{(y-3)(3-x)} = 1$  
That was my attempt, I can't think of anything else here. I'd prefer a hint
 A: Reorder the equation as a second degree equation in $x$ and search for solutions $x$. You find that the discriminant $\Delta$ is always negative for all real value of $y \ne 3$ so there are no real solutions $x$ for real $y \ne 3$ and the solution $x=3$ for $y=3$.

1)Reorder:
$$
x^2-x(3+y)+y^2+9-3y=0
$$
so: $a=1$, $b=-(3+y)$, $c=y^2+9-3y$.
2)Find $\Delta$:
$$
\Delta=b^2-4ac= (3+y)^2-4(y^2+9-3y)=-3y^2+18y-27=-3(y-3)^2
$$
3) see: $\Delta <0 \quad \forall y \in \mathbb{R}-\{3\}$ because $(y-3)^2$ is always positive (for $y \ne 3$) being a sqare.
so there are no $x$ solutions (in $\mathbb{R}$) for all $y \ne 3$ and you can find the solution $x=3$ for $y=3$.
As an exercise do the same reordering for an equation in $y$ .... :)

If you done the exercise you see that all is the same. Why? Because the equation is symmetric for a change of $x$ and $y$, so every solution (if exists) is of the form $x=y=k$. Substituting in the given equation you have the simpler solution:
$2k^2+9=6k+k^2 \iff (k-3)^2=0 \iff k=3$
A: Shift variables to eliminate the linear terms. $x=u+a$ and $y=v+b$, with $a,b$ yet-to-be-determined constants.
$$
\begin{align}
x^2+y^2+9&=3(x+y)+xy\\
(u+a)^2+(v+b)^2+9&=3(u+a+v+b)+(u+a)(v+b)\\
u^2+2au+a^2+v^2+2bv+b^2+9&=3u+3a+3v+3b+uv+av+bu+ab
\end{align}
$$
We would like to be able to cancel the the terms that are linear in $u$ and $v$, so we want:
$$
\left\{\begin{aligned}
2a&=3+b\\
2b&=3+a
\end{aligned}\right.
$$
This system of linear equations in $a,b$ has solution $a=b=3$. So now we have
$$
\begin{align}
u^2+6u+9+v^2+6v+9+9&=3u+9+3v+9+uv+3v+3u+9\\
u^2+v^2&=uv\quad\text{(next, complete the square)}\\
4u^2-4uv+4v^2&=0\quad\text{(multiplying by 4 avoids fractions)}\\
(2u-v)^2+3v^2=0
\end{align}
$$
Since we have two squares adding to $0$, the only solution to this is $u=v=0$, which leads to $x=y=3$.
A: We have to check whether
$$p(x,y):=x^2-xy-3x+y^2-3y+9$$
has real zeros $(x,y)$. This can be done by repeated completing the square:
$$\eqalign{p(x,y)
&=\left(x-{y\over2}-{3\over2}\right)^2-{1\over4}(y^2+6y+9)+y^2-3y+9\cr
&=\left(x-{y\over2}-{3\over2}\right)^2+{3\over4}(y^2-6y+9)\cr
&=\left(x-{y\over2}-{3\over2}\right)^2+{3\over4}(y-3)^2 \ .\cr}$$ 
Here the right hand side is $=0$ iff $y=3$ and $x=3$.
A: Just did this yesterday: What's so special about the form $ax^2+2bxy+cy^2$? 
Given some $g$ that is going to be set equal to $0,$ with
$$ g = a x^2 + b x y + c y^2 + d x + e y + f  $$
and taking the usual
$$  \Delta = b^2 - 4 a c, $$
we get, assuming both $a, \Delta \neq 0$ for this amount of work to be necessary,
$$ \color{red}{ -4a \Delta g} \;  = \; \color{blue}{ (\Delta y + bd - 2ae)^2 - \Delta (2 a x + b y + d)^2} + \color{green}{ \left( \Delta (d^2 - 4 a f) - (bd-2ae)^2 \right)}  $$
I put extra parentheses to emphasize that $\left( \Delta (d^2 - 4 a f) - (bd-2ae)^2 \right)$ is a single constant term, no $x,y$ involved.
New variables to be called:
$$ u = \Delta y + bd - 2ae,  \; \; \; v = 2 a x + b y + d  $$
For you,
$$g= x^2 - xy + y^2 - 3 x - 3 y + 9    $$
is being set to $0,$ and
$$ a=1, b=-1,c=1,d=-3,e=-3,f=9, \Delta = -3,  $$
Go through it all,
$$ -4ag = (-3y+9)^2 + 2 (2x-y-3)^2 $$
and this is being set to $0,$ the constant term came out to $0.$
So we have a single point, $y=3,$ then $2x-y-3 = 0$ says $x=3$ also.
