Since you are talking about rational maps, I assume you mean "open dense" in the Zariski topology, so that $X$ and $Y$ are algebraic varieties. Therefore we have a particular case of the following well-known statement in algebraic geometry. 

> **Proposition.** Let $k$ be an an algebraically closed field of characteristic zero and  $f \colon X \to Y$ a morphism between integral $k$-schemes of finite type over $k$. Then if $f$ is bijective and $Y$ normal the morphism $f$ is an isomorphism, that is $f^{-1} \colon Y \to X$ is a morphism as well.

This is in turn a consequence of the following version of Zariski's main Theorem, see [Görz-Wedhorn, *Algebraic Geometry I*, Theorem 12.83 page 355]. 

> **Theorem.** Let $f \colon X \to Y$ be a separated morphism of finite type such that $f^{\flat} \colon \mathscr{O}_Y \to f_* \mathscr{O}_X$  is an isomorphism. Let $V$ be the open set of $X$ given by the points $x$ such that $\dim f^{-1}(f(x))=0$. Then the restriction $f_{|V} \colon V \to Y$ is an open immersion, hence an isomorphism onto its image. In particular, if $f$ is dominant and all fibres of $f$ are finite then $f$ is birational. 

In fact, given a proper morphism $f \colon X \to Y$ with integral fibres, if $Y$ is normal then $f^{\flat} \colon \mathscr{O}_Y \to f_* \mathscr{O}_X$ is an isomorphism, see again [Görz-Wedhorn, Exercise 12.29 page 365]. 

In your situation, $k= \mathbb{C}$ and $Y$ is smooth (hence normal), because it is an open dense subset of $\mathbb{C}^2$. So the previous results apply. 

See also the following MathOverflow question:
http://mathoverflow.net/questions/73321/isomorphism-between-varieties-of-char-0.