= Solution
At every $z=n\pi$, the inner function has a <simple pole>, with $\cot z=1/(z-n\pi)+h(z)$ for analytic $h$. Thus
$$
e^{\cot z}=e^{h(z)}\exp\left(\frac1{z-n\pi}\right).
$$
The first factor is analytic and nonzero, while the second has infinitely many negative powers in its <Laurent series>. Therefore, by the <exponential of a pole is an essential singularity> result,
$$
\boxed{z=n\pi,\quad n\in\mathbb Z,\quad\text{all essential singularities}.}
$$
At all other finite points the composition is analytic, so there are no additional finite singularities.
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