Answer
$y=-\dfrac{3}{x^3+C}$
Work Step by Step
We are given that $ \dfrac{dy}{dx}=x^2y^2$
We will separate the variables to obtain:
$\dfrac{dy}{y^2}=x^2 \ dx$
Integrate to obtain:
$\int y^{-2} \ dy=\int x^2 \ dx$
This implies that $\dfrac{y^{-1}}{-1}=\dfrac{x^3}{3}+C$
Therefore, we have: $y=-\dfrac{3}{x^3+C}$
