$x^2+y^2+9=3(x+y)+xy$ 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

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.
