The function $$ \exp\left(\tan\left(\frac{1}{z}\right)\right) $$ obviously has a removable singularity at $z=\infty$ and a non-isolated singularity at $z = 0$.
From the real $\tan(x)$ it is clear, that the singularities at $$z_k = \frac{2}{(2k + 1)\pi} \quad (k \in \mathbb{Z})$$ must be essential. Is there an quick and elegant way to show this?
On
The Laurent series of $\tan\left( \frac{1}{z} \right) $ around $z_k$ is $$ \tan\left( \frac{1}{z} \right) = \frac{1/z_k^2}{z - z_k} + h_k(z) $$ where $h_k(z)$ is holomorphic around $z_k$. Then $g$ around $z_k$ takes the form $$ g(z) = \exp\left( \frac{1/z_k^2}{z - z_k} \right) \cdot \exp\left( h_k(z) \right)$$ where the second factor is holomorphic and the first factor clearly has a
Consider $w=1/z$ and $$ \exp(\tan(1/z)) = \exp(\tan(w)) = \exp(\tan(w)) $$ Now set $$ k\pi+\frac{\pi}{2}-w=u $$ so that the expression becomes $$ \exp(\cot u) $$ When $z\to z_k$, we have $w\to 1/z_k$ and $u\to0$. This function has an obvious essential singularity at $0$.