Let $U\subset \mathbb{C}$ be an open set. I know that the function $f:U\rightarrow\mathbb{C}$ takes $f(z_0)\not =0$ and $f''(z_0) \not = 0$ for $z_0\in U$.
How to prove, that $|f|^2$ has no extremum at $z_0$?
Approach: First I write $|f|^2$ as $|f(z)|^2=\Re(z)^2 + \Im(z)^2$. The complex numbers are basically tuples from $\mathbb{R}^2$ so I have to check that $\partial_{\Re} |f|^2 = 2\Re(z) \neq 0$ or $\partial_{\Im} |f|^2 = \Im(z) \neq 0$. Since $f(z_0) \neq 0 \Rightarrow \Re(z_0) \neq 0$. Is this reasoning correct or is that not the way to go?
Since you mention $f''(z_0)\neq0$, we can deduce your function is differentiable and therefore analytic. As a result, we know $|f(z)|$ doesn't have a maximum in $U$ by the maximum modulus principle. Since $|f|$ doesn't obtain a maximum in $U$, neither can $|f|^2$ obtain a maximum.