Let $f : S \to X$ be a dominant morphism of smooth complex compact surfaces. Let $C \subset S$ be a smooth curve such that $df$, seen as a map from $T\_S \to T\_X$, is generically of rank $1$ along $C$ and that in a neighborhood of $C$, $df\_x$ is an isomorphism outside $C$. Suppose that $C$ is not contracted by $f$, is it true that $f\_{|C}$ is an immersion?


One might want to decompose it into two sub-questions:


1. Can one show that the rank of $df$ is exactly $1$ along $C$?
2. Suppose the rank of $df$ is exactly $1$ along $C$, can we show that $f\_{|C}$ is an immersion?



Branch locus along a smooth curve
The question in essentially local on $X$, so the compacteness assumption is irrelevant. 
The answer to both questions is no: consider $S\subset \mathbb C^3$ defined by $z^3+xz+y=0$ and the map to $X=\mathbb C^2$ given by $(x,y,z)\mapsto (x,y)$. 


The ramification curve $C$ is smooth (given by $z^3+xz+y=3z^2+x=0$) while its image has a cusp at $x=y=0$. The differential of the map has rank $\ge 1$ everywhere


