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

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

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

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

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

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

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

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

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*

Excuse me sir isnt it supposed to be 2(y^2)(x) when integrating from the right side

Please indicate which line of the solution you are referring to for us to locate your concern.

$\dfrac{dy}{dx} = \dfrac{x^2 + y^2}{y^3 - 2xy}$ ← Line (1)

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

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

$y^3 \, dy - x^2 \, dx = x(2y \, dy) + y^2 \, dx$ ← Line (4)

$y^3 \, dy - x^2 \, dx = x(2y \, dy) + y^2 \, dx$ ← Line (5)

$y^3 \, dy - x^2 \, dx = d(xy^2)$ ← Line (6)

$\displaystyle \int y^3 \, dy - \int x^2 \, dx = \int d(xy^2)$ ← Line (7)

Im quite confused at line 6 sir when integrating at the right sside

Integrating from the right side at line 6 sir isnt it suppose to be (x)(y^2)+(x)(y)