Solution
= Solution
First retain the <equilibrium> solution $y=0$. For $y\ne0$ and $x\ne\pm1$, <separation of variables> gives $\int y^{-2}dy=\int x/(1-x^2)dx$, hence $-1/y=-\tfrac12\ln|1-x^2|+C$. Renaming the constant yields
$$
\boxed{y=0,\qquad\text{or}\qquad y=\frac2{\ln|1-x^2|+C}.}
$$
A nonzero solution is considered on a regular interval avoiding $x=\pm1$ and zeros of its denominator. The zero solution itself extends through both singular points of the divided equation.