find the solution of the ivp
y' = (1-y^2)x^2
y(0) = 0
The equation is seperatable
$\displaystyle \frac{dy}{dx}=(1-y^2)x^2 \iff \frac{dy}{1-y^2}=x^2dx$
All you need to do from here is integrate.
Note that
$\displaystyle \frac{1}{1-y^2}=\frac{\frac{1}{2}(1-y)+\frac{1}{2}(1+y)}{(1-y)(1+y)}+\frac{\frac{1}{2}}{1+y}+\frac{\frac{1}{2} }{1-y}$