Let $z_0$ be a non-removable isolated singularity of $f$. Show that $z_0$ is then an essential singularity of $\exp(f)$.
Hello, unfortunately I do not know how to proof that. To my opinion one has to consider two cases:
$z_0$ is a pole of order $k$ of $f$.
$z_0$ is an essential singularity of $f$.
On
The above proof is very nice, here I provide another idea. The real question is that if $f$ has a pole, then $e^f$ has an essential singularity.
Assume that the conclusion is false, then there exist an integer $k$ and an analytic function $g$ in some disc $|z|<\delta$, for example, $g(0)\neq0$, such that $e^f=z^kg$,
$$f'=\frac{(e^f)'}{e^f}=\frac{k}{z}+\frac{g'}{g}$$
integrating it on the contour $|z|=\frac{\delta}{2}$ we find $k=0$, since $g$ is analytic. Using the fact that $g(0)\neq0$ we can find another analytic function $h$ in the disc $|z|<\delta$(actually we should shrink this disc), such that $e^f=g=e^h$, so $f=h+2n\pi i$, $f$ has a removable singularity, contradiction!
We can also look at it from the other side.
If $z_0$ is a removable singularity of $e^f$, then $\lvert e^{f(z)}\rvert < K$ in some punctured neighbourhood of $z_0$. Since $\lvert e^w\rvert = e^{\operatorname{Re} w}$, that means $\operatorname{Re} f(z) < K'\; (= \log K)$ in a punctured neighbourhood $\dot{D}_\varepsilon(z_0)$ of $z_0$, and that implies that $z_0$ is a removable singularity of $f$. (Were it a pole, $f(\dot{D}_\varepsilon(z_0))$ would contain the complement of some disk $D_r(0)$; were it an essential singularity, each $f(\dot{D}_\varepsilon(z_0))$ would be dense in $\mathbb{C}$ by Casorati-Weierstraß; in both cases $\operatorname{Re} f(z)$ is unbounded on $\dot{D}_\varepsilon(z_0)$.)
If $z_0$ were a pole of $e^f$, it would be a removable singularity of $e^{-f}$, hence $z_0$ would be a removable singularity of $-f$ by the above, hence $z_0$ would be a removable singularity of $f$, and therefore a removable singularity of $e^f$ - contradiction.