Holomorphic diffeomorphism which inverse is not holomorphic? Is there such a thing as a (germ at the origin of) holomorphic mapping  $f:\mathbb{C}^n\rightarrow\mathbb{C}^n$ which admits a differentiable inverse (of any class) which is not holomorphic?
The motivation for this question is that I'm seeing in some publications the expression "holomorphic diffeomorphism" as equivalent (or being used to refer to) biholomorphisms. My impression is that the word "diffeomorphism" is given emphasis due to the fact that it represents a new system of coordinates in $\mathbb{C}^n$.
This all matters only locally (at the origin) to what I'm studying, so I included the possibility for germs only, but I wondered if that was true in general as well.
 A: Here is a sketch, hopefully all correct. (I think I'm answering your question.) What do you think?
Prop: A smooth function $f : U \subset \mathbb{C}^n \to \mathbb{C}$ is holomorphic iff $\partial f / \partial \bar{z} = 0$  (constantly zero). Pf:This is equivalent to the CR equations. Here $\partial f/ \partial \bar{z}$ means the vector $(\partial f / \partial \bar{z_i})$, so this is equivalent to CR in each coordinate direction.
For $f : \mathbb{C}^n \to \mathbb{C}^m$, define $D(f) = (\frac{ \partial f_i}{\partial z_j})$ and $\bar{D}(f) =  (\frac{ \partial f_i}{\partial \bar{z}_j})$. 
Prop: Then $f$ is holomorphic iff $\bar{D}(f) = 0$. Pf: This follows from above.
Prop: (Chain rule) Suppose that $U \subset \mathbb{C}^n$, $V \subset \mathbb{C}^m$ are open sets, and $f : U \to V$ is smooth, and $g : V \to \mathbb{C}^k$ is smooth. Then $D(g \circ f) = D(g) \bar{D}(f) + \bar{D}(g) \bar{D}(\bar{f})$
Pf: Long computation in coordinates. (This is another way to write 1.3.1 in the linked notes - this is the spot I would check my writing carefully.)
Prop: $\bar{D}(\bar{f}) = \overline{ D(f)}$. Pf: The point is that $\overline{ (\partial_x + i \partial_y)(u + iv)} = \overline{ (\partial_x + i \partial_y)} \dot{} \overline{(u + iv)}$. 
Prop: Suppose that $f$ is holomorphic, and $g \circ f$ is holomorphic. Suppose that $D(f)$ is everywhere invertible . Then $g$ is holomorphic.
$0 = \bar{D}(gf) = D(g) \bar{D}(f) + \bar{D}(g) \bar{D}({\bar{f}}) = \bar{D}(g) \overline{ D(f) }. $
Since the last term is an invertible matrix, this implies that $\bar{D}(g) = 0$. Hence $g$ is holomorphic.
