Note that $2x^3y + xy^2 = x^2+xy+3x^2y$. We can subtract this from the first equation to deduce $2x^3y = 3x^2y+xy-y$. So then $y(2x^3-3x^2-x+1)=0$ or $y(2x-1)(x^2-x-1)=0$. Case 1: $y=0$. Then $x=0$ by equation 1 and this works. Case 2: $x=\frac{1}{2}$. Then $2y^2-4y-1=0$ by equation 1 and you can check these. Case 3: $x^2=x+1$. Plug this into equatio 2 and it reduces nicely to $y+y^2=x+xy$ or $(y+1)(y-x)=0$ and you can check that these work..