Notice that
[tex](2x^3-xy^2-2y+3)_y=-2xy-2[/tex]
[tex](-x^2y-2x)_x=-2xy-2[/tex]
so the ODE is exact, and we can find a solution [tex]F(x,y)=C[/tex] such that
[tex]F_x=2x^3-xy^2-2y+3[/tex]
[tex]F_y=-x^2y-2x[/tex]
Integrating both sides of the first equation wrt [tex]x[/tex] gives
[tex]F(x,y)=\dfrac{x^4}2-\dfrac{x^2y^2}2-2xy+3x+g(y)[/tex]
Differentiating wrt [tex]y[/tex] gives
[tex]F_y=-x^2y-2x=-x^2y-2x+g'(y)\implies g'(y)=0\implies g(y)=C[/tex]
So we have
[tex]\boxed{F(x,y)=\dfrac{x^4}2-\dfrac{x^2y^2}2-2xy+3x=C}[/tex]
