<s>[INST] $x^2+y^2+9=3(x+y)+xy$ 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
 [/INST] 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 $x=y=3$.
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