How to solve $y'' + (1-y)(1+y)y = 0$?

$$y''+(1-y)(1+y)y =0$$ Swithch variables to get $$-\frac{x''}{[x']^3}+(1-y)(1+y)y=0$$ Reduction of order $p=x'$ gives $$p=x'=\pm \frac 1 {\sqrt{\frac{y^4}{2}-y^2+c_1 }}$$ and now, you will face an ugly ellptic integral. $$x+c_2=\pm \int \frac {dy} {\sqrt{\frac{y^4}{2}-y^2+c_1 }}$$

Rewrite $$\sqrt{\frac{y^4}{2}-y^2+c_1 }=\frac 1 {\sqrt 2} \sqrt{(y^2-a)(y^2-b)}$$ where $a$ and $b$ are the roots of the quadratic in $y^2$ and, hoping that I am not mistaken, $$x+c_2=\pm \sqrt{\frac{2}{b}}F\left(\sin ^{-1}\left(\frac{y}{\sqrt{a}}\right)|\frac{a}{b}\right)$$

Do not try to inverse it (except if you want the monster @Jean Marie posted).