The question is
Let $f $ be analytical and not constant in $\{z\in\mathbb C:|z|\leq 1\}$. If there is some $z_0\in \{|z|\leq 1\}$ such that $$|f(z_0)|=\max_{|z|\leq 1} |f(z)|\,,$$ then there must be $|z_0|=1 $ and $f’(z_0)\neq 0$.
The first part is just the maximum principle, but for the derivative, I have no idea how solve it. Looking forward to some hints or answers, thank you.
The hypothesis means that $f$ has an analytic extension to a neighborhood of $z_0$ (actually it means that $f$ is analytic on a slightly larger open disc, but we need only the weaker assumption above)
$f$ nonconstant means that $f(z_0) \ne 0$ and $|f(re^{it}z_0)| < |f(z_0)|, 0 \le r <1$. To simplify the proof, by a rotation and homothety let wlog $f(z_0)=1$ (take $g(z)=Af(\alpha z), |\alpha| =1, A>0$ as that doesn't change the problem).
Assuming $f'(z_0)=0$ we have a power series $$f(z)=1+a(z-z_0)^n+O(|z-z_0)|^{n+1}, a \ne 0, n \ge 2,$$ for $|z-z_0|$ small enough.
But now if $\arg a =\theta$, since $n \ge 2, n\arg (z-z_0)$ covers more that an arc of angle length $\pi$ when $z$ is close to $z_0$ and $|z|<1$ so we can find such for which $\theta+n\arg (z-z_0) \in (-\pi/2, \pi/2)$ modulo $2\pi$ which means that $\Re a(z-z_0)^n >0$.
Letting $z \to z_0$ radially we can insure that $\Re (a(z-z_0)^n+O(|z-z_0)|^{n+1}) >0$ as the remainder decays faster, so one can find $|w|<1, \Re (a(w-z_0)^n+O(|w-z_0)|^{n+1}) >0$ which means that $|f(w)| \ge \Re f(w) >1=|f(z_0)|$ and that is a contradiction!