I'm reviewing old homework problems.
Let a function $f(z)$ be some branch of $z^{1/n}$. Show that $$f'(z) = \frac{f(z)}{nz} \textbf{.}$$
I wrote:
Let $z = re^{i(\theta+2\pi k)}$ and fix $k$ thus for $f(z) = z^{1/n}$, the branch is fixed. Since $f(z)$ is analytic everywhere, we can differentiate thus $f'(z) = \frac{1}{n} z^{1/n \, -1}\, {\color{red} =} \,\frac{1}{n} \frac{z^{1/n}}{z}$, for a fixed branch.
On my homework, the professor pointed at the equal sign that I've marked in red and said: both sides are multivalued so it is unclear what this line means.
So I don't know how I would go about this problem. In general, branch cuts confuse me. I understand that the function $f(z) = z^{1/n}$ has different branches and that we need to choose one branch to allow the function to be continuous and single valued.