Answer
$ y = -1/(x^3 + C) $
Work Step by Step
$dy/dx = 3x^2y^2$
Let's separate the y on one side and the x on the other side:
$ (1/y^2 )dy = 3x^2dx$
Now let's integrate both sides:
$\int (1/y^2 )dy = \int 3x^2dx$
$-1/y = x^3 + C$
Isolate y:
$ y = -1/(x^3 + C) $
