In this question, I recently asked on if the isolated singularities of a meromorphic function $f$ are the same ones as the of the function $e^f$. I quickly realized that this isn't the case, so I hereby wanted to rephrase the question more generally:
Given a meromorphic function $f: U \to \mathbb{C}, U \subseteq \mathbb{C}$ open, and an isolated singularity $a \in U$ of $f$, does the kind of singularity (removable, pole or essential) that $a$ is for $f$ already determine the kind of singularity that $a$ is for the function $g(z) := e^{f(z)}$? (Notice that, unlike in the other question, I'm not demanding that they are the same kind of singularity – just if we already know which kind of singularity $a$ is for $e^f$ if we know the kind regarding $f$.)
I unfortunately don't have much more to offer as an approach as I did in the other question. I know that if $a$ is a removable singularity for $f$, then it's a removable singularity for $e^f$ aswell. And for many functions with a pole, $e^f$ has an essential singularity, with one of the most basic examples being $f(z) = \frac{1}{z}$ (as it was also given as an answer in the linked question).
So I'm basically searching for any functions $f$ with a pole $a$, so that $e^f$ actually also has a pole at $a$ (since I don't expect a pole to become a removable singularity). Or if there is a function $f$ with an essential singularity $a$, so that $e^f$ has a pole. (Since I believe that most essential singularities are sent to essential singularities, if we consider for example $f(z) = e^{\frac{1}{z}}$. I would then expect $e^{e^{\frac{1}{z}}}$ to have an essential singularity aswell.)
If I'm wrong though and the kind of singularity of $f$ does indeed determine the kind of singularity of $e^f$... then I would appreciate any hints or approaches on how to actually proof that; because so far, I've been out of ideas.