$\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{y^3 - 2xy}$

$(y^3 - 2xy) \, dy = (x^2 + y^2) \, dx$

$y^3 \, dy - 2xy \, dy = x^2 \, dx + y^2 \, dx$

$y^3 \, dy - x^2 \, dx = x(2y \, dy) + y^2 \, dx$

$y^3 \, dy - x^2 \, dx = x(2y \, dy) + y^2 \, dx$

$y^3 \, dy - x^2 \, dx = d(xy^2)$

$\displaystyle \int y^3 \, dy - \int x^2 \, dx = \int d(xy^2)$

$\frac{1}{4}y^4 - \frac{1}{3}x^3 = xy^2 + C$

At (3, -2)

$\frac{1}{4}(-2)^4 - \frac{1}{3}(3^3) = 3(-2)^2 + C$

$C = -17$

Thus,

$\frac{1}{4}y^4 - \frac{1}{3}x^3 = xy^2 - 17$

$3y^4 - 4x^3 = 12xy^2 - 204$ *answer*