= Solution
Away from zeros of the leading coefficient, the normal-form <differential equation> is
$$
y''+P(z)y'+Q(z)y=0,\qquad P(z)=\frac3{z(1+z^2)^2},\qquad Q(z)=\frac1{z^2(1+z^2)^2}.
$$
The only finite singular points are $z=0,i,-i$. At zero, $zP(z)=3/(1+z^2)^2$ and $z^2Q(z)=1/(1+z^2)^2$ are analytic, so zero is a <regular singular point>. At $z=\pm i$, the coefficient $P$ has a genuine double <pole>, so $(z\mp i)P$ is not analytic; these are <irregular singular points>. Thus
$$
\boxed{z=0\text{ regular singular};\qquad z=\pm i\text{ irregular singular}.}
$$
If infinity is included, set $t=1/z$ and $Y(t)=y(1/t)$. The transformed coefficients are $\widetilde P=2/t-3t^3/(1+t^2)^2$ and $\widetilde Q=t^2/(1+t^2)^2$. They satisfy the <regular singular point at infinity> criterion, so $\boxed{\infty\text{ is regular singular}}$ as well.
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