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