$y' = xy^2$
$\dfrac{dy}{dx} = xy^2$
$\dfrac{dy}{y^2} = x \, dx$
$\displaystyle \int y^{-2} \, dy = \int x \, dx$
$\dfrac{y^{-1}}{-1} = \dfrac{x^2}{2} + \dfrac{c}{2}$
$-\dfrac{1}{y} = \dfrac{x^2}{2} + \dfrac{c}{2}$
$-2 = x^2y + cy$
$x^2y + cy + 2 = 0$
$y(x^2 + c) + 2 = 0$ answer