I'm trying to understand a proof of the title's statement. The proof is as follows:
If $f(z)$ is an elliptic function (doubly periodic and meromorphic (of finite order), $f'(z)/f(z)$ is also doubly periodic and is meromorphic because: suppose $f(z)= (z-w_i)^{m_i} h(z)$ where $h(z)$ is holomorphic. Then we have: $f'(z)/f(z)= \frac{m_i}{(z-w_i)} +h'(z)/h(z).$
at this point of proof, it is said that $h'(z)/h(z)$ is holomorphic and I don't know why. (please explain it why)
Then we use the Cauchy residue integral theorem and conclude that the sum of $m_i$ has to be zero and we are done.
The point is that $h(z)$ is holomorphic and nonzero at $z=w_i$, so that $h'/h$ is holomorphic in a neighbourhood of $w_i$, so the residue of $f'/f$ at $w_i$ is $m_i$.