Equation of a curve with given equation of slope and passing through a point

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

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

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

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

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

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

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

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