= Solution
{c}
Away from zero, singularities occur exactly when $e^{1/z}=-1$, namely
$$
\boxed{z_k=\frac1{(2k+1)\pi i},\qquad k\in\mathbb Z.}
$$
For $D(z)=z(e^{1/z}+1)$, differentiation at $z_k$ gives $D'(z_k)=1/z_k\ne0$. Thus every $z_k$ is a simple <pole>, with <residue> $z_k$.
As $|k|\to\infty$, these <poles> accumulate at zero. Therefore \b[zero is a <non-isolated singularity>]. Every punctured disc around it contains poles, so the function is not holomorphic on any punctured disc there. In particular zero cannot be classified as an isolated <essential singularity>; the isolated-singularity classification does not apply. The displayed poles and their <accumulation point> zero exhaust the finite singularities.
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