Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$ 
Find all integer solutions $(x,y)$ such that $2x^2 + y^2 = 3x^2y$.

We can rearrange the given equation to $$y^2 = x^2(3y-2)\tag1$$ Thus $3y-2$ must be a perfect square and so $3y-2 = k^2$.
How can we continue?

$x=0\iff y=0.$ And $x=y=0$ is a solution.
If $x\ne 0\ne y$ then $$2x^2+y^2=3x^2y\implies x^2(2-3y^2)=-y^2\ne 0.$$  This implies $x^2$ divides $y^2.$ Now $y^2/x^2$ is an integer for non-zero integers $x,y$ only if  $x$ divides $y$.  So we have $$y=zx$$ with  integer $z.$
So $2x^2+z^2x^2=3x^3z$. Since $x\ne 0$ we have $2+z^2=3xz.$... So $$z(z-3x)=-2$$ which implies that $z$ divides $2$.
Putting $y=zx$ with $z\in \{\pm 1 ,\pm 2\}$ into the original equation gives us the non-zero solutions :$$(y=1, x=\pm 1), (y=2,x=\pm 1).$$ For example if $z=1$ then $y=x$ and $3x^2=2x^2+y^2=3x^2y=3x^3\ne 0$ implies $1=x=y.$
