exp(1/z) has an essential singularity at 0

Remarks

A direct check: for real $x \to 0^+$, $e^{1/x} \to \infty$, and for real $x \to 0^-$, $e^{1/x} \to 0$. So $f$ has no limit at $0$ and $|f|$ does not tend to $\infty$. In each punctured disc $0 < |z| < \delta$, $f$ takes each value $w \neq 0$ infinitely often: the solutions of $e^{1/z} = w$ are $z = 1 / (\operatorname{Log} w + 2\pi i k)$ with $k \in \mathbb{Z}$ and $\operatorname{Log} w + 2\pi i k \neq 0$, and these points tend to $0$ as $|k| \to \infty$. The value $0$ is never taken. This agrees with Picard's great theorem, which allows one exception. The residue of $f$ at $0$ is the coefficient of $z^{-1}$, which is $1$.