Since you are talking about rational maps, I assume you mean "open dense" in the Zariski topology, so that your $X$ and $Y$ are algebraic varieties. Then this is 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.
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 result applies.
See also the following MathOverflow question: Isomorphism between varieties of char 0