 Solution

Away from zeros of the leading coefficient, the normal-form differential equation isThe only finite singular points are $z=0,i,-i$. At zero, $zP(z)=3/(1+z^2{)}^2$ and $z^{2}Q(z)=1/(1+z^2{)}^2$ 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. ThusIf infinity is included, set $t=1/z$ and $Y(t)=y(1/t)$. The transformed coefficients are $\tilde{P}=2/t-3t^3/(1+t^2{)}^2$ and $\tilde{Q}=t^2/(1+t^2{)}^2$. They satisfy the regular singular point at infinity criterion, so $\boxed{\infty \text{ is regular singular}}$ as well.