Solve $$x^3y^2(2y-x)=x^2y^4-36$$ where $x$ and $y$ are integers.
Preparing for Math Olympiads and I am totally unable to come up with a method of solving this equation without factoring (if there is one) and with factoring I don't know how to factor it because every pair of $xy$ has varying degrees on $x$ and $y$ that make it complicated to factor. Are there any methods to solving/factoring questions like these and what should be my first thought process. Thank you.
Hint:
Observe, $x$ and $y$ are non-zero integers.
$x^2y^2$ divides $36\implies 36=x^2y^2\cdot k,$ for some positive integer $k.$ So, $$x(2y-x)=y^2-k\implies (x-y)^2=k\implies xy(x-y)=\pm6$$
Now, list the divisors of $6$ and some case work should finish it off.