= Solution
In normalized form, the coefficients are $p(x)=-x/(1-x^2)$ and $q(x)=n^2/(1-x^2)$. They are analytic near zero, so $\boxed{x=0\text{ is an ordinary point}}$.
At each of $x=\pm1$, $p$ has a simple pole and $q$ at most a simple pole. The products $(x-x_0)p$ and $(x-x_0)^2q$ are analytic, so both are <regular singular points>. More precisely their limiting values are $1/2$ and zero, giving <indicial equation> $\rho(\rho-1)+\rho/2=0$, with exponents $0,1/2$.
To classify infinity, set $t=1/x$ and $Y(t)=y(1/t)$. The transformed equation is
$$
t^2(t^2-1)Y''+(2t^3-t)Y'+n^2Y=0.
$$
After division, its coefficients satisfy
$$
tP(t)=\frac{2t^2-1}{t^2-1}\longrightarrow1,\qquad t^2Q(t)=\frac{n^2}{t^2-1}\longrightarrow-n^2.
$$
Both are analytic at $t=0$, while $P$ itself has a pole. Hence $\boxed{x=\infty\text{ is also a regular singular point}}$, with indicial exponents $\rho=\pm n$. These classifications remain valid at $n=0$, when the exponents at infinity coincide.
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