Let $\pi:S\to C$ be a holomorphic map, where $S$ is a complex surface and $C$ is a complex curve. For a fiber $\pi^{-1}(c) $ of $\pi$, suppose there is a curve $F$ in $\pi^{-1}(c)$ such that the differential $d\pi$ vanishes along $F$. Then for a smooth point $p\in F$, how can we show that $\pi$ is locally (near $p$) given by $\pi(z_1,z_2)=z_1^m$ for some integer $m\geq 2$?
My try: Since $p$ is a smooth point of $F$, $F$ is embedded in $S$ near $p$, so we can choose a chart neighborhood $U$ near $p$ so that $U\cap F=\{(z_1,z_2):z_1=0\}$ and $p=(0,0)$. Now $\pi$ locally satisfies $\pi(0,z_2)=0$. Maybe now I should use $d\pi$ vanishes along the line $z_1=0$ but I got stuck here.
I might have misunderstood your condition but it seems that $p$ may be the intersection of several degenerate curves. Take e.g. $\pi(z_1,z_2)=z_1^m z_2^n$. The condition of $p$ being smooth in $F=\{z_1=0\}$ does not guarantee that $F$ is the only zero locus curve going through $p=(0,0)$.