= Solution
Away from $y=\pm1$, divide by the coefficient of $x''$ to obtain the normal form
$$
x''+P(y)x'+Q(y)x=0,\qquad P(y)=Q(y)=\frac1{y^2-1}.
$$
Thus $\boxed{y=1\text{ and }y=-1}$ are the only finite singular points. For a second-order <linear differential equation>, $y_0$ is a <regular singular point> if $(y-y_0)P(y)$ and $(y-y_0)^2Q(y)$ extend analytically to $y_0$. At $y_0=1$ these functions are
$$
\frac1{y+1},\qquad \frac{y-1}{y+1},
$$
and at $y_0=-1$ they are
$$
\frac1{y-1},\qquad \frac{y+1}{y-1}.
$$
Each pair is analytic at its respective point. Therefore $\boxed{\text{both finite singular points are regular}}$.
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