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?

We note that $x^2\mid 3x^2y$, so $x^2\mid 2x^2+y^2$, then $x^2\mid y^2$, which implies that $x\mid y$. Let's call $y=kx$, if we replace this into your equation $(1)$ we get $k^2x^2=x^2(3kx-2)$, then $k^2=3kx-2$, which lead us to the quadratic equation $k^2-3xk+2=0$. Using the formula for quadratic equations we deduce that $$k=\frac{3x\pm \sqrt{9x^2-8}}{2}.$$
So in order to have $k\in\mathbb{Z}$ we must have $9x^2-8=z^2$, for some $z\in  \mathbb{Z}$, so $(3x+z)(3x-z)=8$, therefore $3x+z$ and $3x-z$ are both divisors of $8$ with the same parity. Hence $3x+z=\pm 4$ and $3x-z=\pm 2$, or $3x+z=\pm 2$ and $3x-z=\pm 4$. 
For example if $3x+z=4$ and $3x-z=2$, then $x=1$ and $z=1$, which gives us $k=1$ or $k=2$, and thus $y=1$ or $y=2$. The other cases are similar.
