Yes. First, smoothness implies that $k(X)$ is a separable field extension of $k$, cf. Corollary 16.17 of Eisenbud's "Commutative Algebra". Thus there exists an element $f\in k(X)$ such that $k(X)$ is a finite, separable extension of the subfield $k(f)$. Considering $f$ as a rational function on $X$, $f$ extends to a regular $k$-morphism $f:X\to \mathbb{P}^1_k$ which is generically \'etale.
– Jason StarrMay 1 '12 at 12:01

I'm 3 minutes late. The only thing I can add is that, in case you don't have Eisenbud at hand, but you have Lang's "Algebra", then the reference for infinite separable extensions is Prop VIII.4.1.
– JefMay 1 '12 at 12:08

Thank you for answering !
– kisekiMay 1 '12 at 13:14