So I am stuck in where he says clearly at the end.
Proposition 1.10 (Implicit function theorem) Let $U \subset \mathbb{C}^m$ be an open subset and let $f:U \rightarrow \mathbb{C}^n$ be a holomorphic map where $m \ge n$. Suppose $z_0 \in U$ is a point such that $$\text{det}\bigg(\frac{\partial f_i}{\partial z_j}(z_0) \bigg)_{1 \le i,j \le n} \neq 0.$$ Then there exists open subsets $U_1 \subset \mathbb{C}^{m-n}$, $U_2 \subset \mathbb{C}^n$ and a holomorphic map $g:U_1 \rightarrow U_2$ such that $U_1 \times U_2 \subset U$ and $f(z)=f(z_0)$ if and only if $$g(z_{n+1},...,z_m)=(z_1,...,z_n)$$ Pf. Using the relation between the complex and real Jacobian explained above, one finds $z$ is regular if and only if $$\text{det}J_\mathbb{R}(z)\neq 0.$$ That is, $z$ is a regular point for the underlying real map. Thus the real inverse theorem applies and we are guaranteed the existence of a $C^\infty$ function $f^{-1}:V' \rightarrow U'$ of $f$. It then suffices to prove $f$ satisfies the C-R equations. Then he says clearly
$$\frac{\partial (f^{-1} \circ f)}{\partial \overline{z}_j}=0$$
Is this because $(f^{-1} \circ f)(z) =z$