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Yes, $f^{-1}$ is holomorphic. In fact, the following result holds, see [Griffiths-Harris, Principle Principles of Algebraic Geometry p. 19].

Proposition

If $f \colon U \to V$ is a one-to-one holomorphic map of open sets of in $\mathbb{C}^n$, then $|J_f| \neq 0$, that is $f^{-1}$ is holomorphic.

The fact that $N$ is smooth is crucial. For instance, if $N \subset \mathbb{C}^2$ is the cuspidal cubic curve of equation $y^2=x^3$ and $f \colon M \to N$ is the normalization map, then $f$ is bijective and holomorphic but it is not a biholomorphism, since $f^{-1}$ is not holomorphic at the point $0$. (0,0)$.

show/hide this revision's text 1

Yes, $f^{-1}$ is holomorphic. In fact, the following result holds, see [Griffiths-Harris, Principle of Algebraic Geometry p. 19].

Proposition

If $f \colon U \to V$ is a one-to-one holomorphic map of open sets of $\mathbb{C}^n$, then $|J_f| \neq 0$, that is $f^{-1}$ is holomorphic.

The fact that $N$ is smooth is crucial. For instance, if $N \subset \mathbb{C}^2$ is the cuspidal cubic curve of equation $y^2=x^3$ and $f \colon M \to N$ is the normalization map, then $f$ is bijective and holomorphic but it is not a biholomorphism, since $f^{-1}$ is not holomorphic at $0$.