1.

After multiplying the equation by y, we have M = 2 x y³ - y²and N = 3 x² y² - 2 x y + 2 y. M_y = 6 x y² - 2 y = N_xso this makes the DE exact, i.e. y is an integrating factor of the original DE. $F(x,y) = \int M \, dx = x^2 y^3 - x y^2 + h(y)$.

$$N = 3 x^2 y^2 - 2 x y + 2 y = \frac{\partial F}{\partial y} = 3 x^2 y^2 - 2 x y + h'(y)$$

So h'(y) = 2 y. We take h(y) = y², so the solutions are x² y³ - x y² + y² = constant.