Let $f:\mathbb{C}\to\mathbb{C}$ be an analytic function on $z=0$ and let $g:\mathbb{C}\to\mathbb{C}$ such that $g(z)=e^{\frac{1}{z}}f(z)$. Given $f(z)\not\equiv0$, prove that $z=0$ is an essential singularity of $g(z)$.
The only way I know to prove a point $z_0$ is an essential singularity of a function $g(z)$, is to prove that the Laurent Series of the function $g(z)$ around the point $z_0$ has an infinite amount of negative order powers. So this is what I tried to do.
First, if $f(z)\not\equiv0$, then $z=0$ is a zero of a finite order of $f(z)$; meaning there exists $m\in\mathbb{N}$ such that $f(z)=z^mh(z)$, when $h(z)$ is analytic on $z=0$ and $h(z)\neq0$. Then:
$$g(z)=e^\frac{1}{z}z^mh(z)$$
If $h(z)$ is analytic on $z=0$, then it has a Taylor Series around it: $h(z)=\sum_{n=0}^{\infty}h_nz^n$ when actually $h_n=\frac{h^{(n)}(0)}{n!}$ for every $n\in\mathbb{N}$. To conclude:
$$g(z)=\sum_{k=0}^{\infty}\frac{1}{k!z^k}\sum_{n=0}^{\infty}h_nz^{n+m}$$
This unfortunately doesn't promise that the actual Laurent Series of $g(z)$ would have an infinite amount of negative order powers (since they can cancel each other, allegedly).