= Solution
First simplify the coefficient of $y$ using
$$
\frac{m^3}{x-m}+x^2+mx+m^2=\frac{x^3}{x-m}.
$$
After dividing by $x(x-m)$, the <second-order linear differential equation> is
$$
y''+P(x)y'+Q(x)y=0,\qquad P(x)=\frac1x+\frac1{x-m},\qquad Q(x)=\frac{\omega^2x^2}{(x-m)^2}.
$$
At zero, $xP\to1$ and $x^2Q\to0$, and both products are analytic there. At $m$, $(x-m)P\to1$ and $(x-m)^2Q\to m^2\omega^2$, again with analytic extensions. Thus $\boxed{x=0\text{ and }x=m\text{ are regular singular points}}$, and there are no other finite singular points.
Infinity must also be considered because the question asks for all singular points. Put $t=1/x$ and $Y(t)=y(1/t)$. The coefficient of $Y$ in its normal-form equation is $Q(1/t)/t^4\sim\omega^2/t^4$. Multiplication by $t^2$ does not make it analytic at zero. Hence $\boxed{x=\infty\text{ is an irregular singular point}}$.
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