First, I can show that $f$ meromorphic is a rational function. Now, I want to consider $g=e^{f(z)}$.
I have heard that there is something interesting that goes on with $g$, that there is some room for interpretation.
Of course, the quick answer would be "no, where $f$ has a pole, $g$ would have infinitely many negative power terms in $(z-z_0)$ in its Laurent expansion, and by definition $g$ has an essentially singularity there and hence cannot be meromorphic."
Are there conditions that would make $g$ meromorphic? Even entire?
$f$ is only assumed to be meromorphic but not entire.
Thanks,
Suppose $f$ has a pole at a point $a$. Then the function $h(z)=1/f(z)$ is holomorphic in a neighborhood of $a$ with $h(a)=0$. By the open mapping theorem, if $B$ is any ball around $a$, then $h(B)$ contains a ball around $0$. It follows that there exists an $R>0$ such that $f(z)$ takes all values of modulus $>R$ on $B$. In particular, by the $2\pi i$-periodicity of $\exp$, $g(z)=\exp(f(z))$ takes every value other than $0$ on $B$ (since for any nonzero value, we can find a logarithm of it with imaginary part $>R$).
That is, $g$ takes every value except $0$ in every neighborhood of $a$. It follows that $a$ must be an essential singularity of $g$. So if $f$ has poles, $g$ must have essential singularities, and hence cannot be meromorphic.