How exactly does writing $f(z)=f(z,\bar{z})$ work? I know that we can just define the differential operators
$$\frac{\partial}{\partial z} = \frac{1}{2}(\frac{\partial}{\partial x} - i \frac{\partial}{\partial y})$$
$$\frac{\partial}{\partial \bar{z}} = \frac{1}{2}(\frac{\partial}{\partial x} + i \frac{\partial}{\partial y})$$
and that $f$ is holomorphic means that $\frac{\partial f}{\partial \bar{z}}=0$, that's all clear. However, what I don't get it is how we can write any arbitrary function $f(z)=f(z,\bar{z})$, and then calculate 'formally' with the above differential operators and expect the outcomes to be as expected (or even well defined). 
So this question is really not about the derivation/definition of the  Wirtinger Derivatives, that's all document very well. It's about why we can write for example
\begin{align}
\frac{\partial (g\circ f)}{\partial z}&=\frac{\partial (g(f(z,\bar z),\bar f(z,\bar z))}{\partial z}\\\\
&=\left.\frac{\partial g(w,\bar w)}{\partial w}\right|_{w=f(z,\bar z)}\times \frac{\partial f(z,\bar z)}{\partial z}+\left.\frac{\partial g(w,\bar w)}{\partial \bar w}\right|_{\bar w=\bar f(z,\bar z)}\times \frac{\partial \bar f(z,\bar z)}{\partial z}\\\\
&=\left(\frac{\partial g}{\partial z}\circ f\right)\frac{\partial f}{\partial z}+\left(\frac{\partial g}{\partial \bar z}\circ f\right)\frac{\partial \bar f}{\partial z}
\end{align}
(copied from here). I just don't see how the formal definition of the Wirtinger derivatives makes it so that all of this goes through.
An answer to this question would explain, rigorously, why the steps in the above calculation are justified, starting from the fact we can write $f(z)=f(z,\bar{z})$ in a well-defined way s.t. the operators $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial \bar{z}}$ behave as expected. 
Thank you
 A: Note that the referenced post does not suggest that $f$ is a function of $z$ only and nowhere do we see written "$f(z)=f(z,\bar z)$."  Rather, a complex-valued function, $f$, is in general, a function of both $z$ and $\bar z$.   To see this, let's take a closer look at things.
Let $\hat f$ be a complex function.  Then we can write $\hat f$ in terms of its real and imaginary parts
$$\hat f(x,y)=u(x,y)+iv(x,y) \tag 1$$
where $ u(x,y)$ and $ v(x,y)$ are real=valued functions of $x$ and $y$ with
$$u(x,y)=\text{Re}(\hat f(x,y))$$
and 
$$v(x,y)=\text{Im}(\hat f(x,y))$$
Next, note that we can write $x$ and $y$ in terms of $z=x+iy$ and $\bar z=x-iy$ as 
$$x=\frac12(z+\bar z) \tag 2$$
and
$$y=\frac{1}{2i}(z-\bar z) \tag 3$$
Substituting $(2)$ and $(3)$ into $(1)$ reveals 
$$\begin{align}
\hat f(x,y)&= u\left(\frac12(z+\bar z),\frac{1}{2i}(z-\bar z)\right)+i v\left(\frac12(z+\bar z),\frac{1}{2i}(z-\bar z)\right)\\\\
&=f(z,\bar z)\\\\
\end{align}$$
for some function $f$ of $z$ and $\bar z$.  So, any complex-valued function that can be expressed as in $(1)$ can be expressed as a function of $z$ and $\bar z$.
A: Background
Throughout, I write things like $f:X\to Y$ when I either mean a partial function $f:X\not\to Y$, or at least that I only care about and am making claims about nice properties on a contextually-appropriate subset of $X$. It wasn't worth giving names to all of the relevant subsets.
For a reference on real and complex series of multiple variables, see Chapter 1 of Notes on global analysis by Andrew D. Lewis. For a higher level approach to the Wirtinger derivatives, see Why Wirtinger derivatives behave so well in the writings of MathSE's own Bart Michels.
Two Input Variables
Suppose we have a complex function $f:\mathbb{C}\to\mathbb{C}$ that is nice enough to have real-analytic components on some open domain $D$, like $f(z)=|z|+\exp(\overline{z})$. Then via $x=\Re z,y=\Im z,u=\Re f,v=\Im f$, we can interpret this as a function $\mathbf{f}:\mathbb{R}^{2}\to\mathbb{R}^{2}$ like $\mathbf{f}(x,y)=\left(u(x,y),v(x,y)\right)$ where $u$ and $v$ are analytic on $D$. Since $D$ is open, the series for $u$ and $v$ are absolutely convergent, so that we can use the power series to extend $u$ and $v$ to complex functions $\widetilde{u},\widetilde{v}:\mathbb{C}^{2}\to\mathbb{C}$. These can be put together to form $\widetilde{\mathbf{f}}:\mathbb{C}^{2}\to\mathbb{C}^{2}$. Inspired by "$x=(z+\overline{z})/2$" and "$y=(z-\overline{z})/(2i)$", we can define a helper function $\mathbf{h}:\mathbb{C}^{2}\to\mathbb{C}^{2}$ given by $h\left(z_{1},z_{2}\right)=\left((z_{1}+z_{2})/2,(z_{1}-z_{2})/(2i)\right)$. Define $\widehat{\mathbf{f}}:\mathbb{C}^{2}\to\mathbb{C}^{2}$ to be the composition $\widetilde{\mathbf{f}}\circ\mathbf{h}$. Note that for any $z\in\mathbb{C}$, we have $\widehat{\mathbf{f}}\left(z,\overline{z}\right)=\left(\Re f(z),\Im f(z)\right)\in\mathbb{R}^{2}$.
Then, at least on a compact subset of $D$ (so that we have uniform convergence for the series), the complex partial derivative $\left.\dfrac{\partial\widehat{\mathbf{f}}(z_{1},z_{2})}{\partial z_{1}}\right|_{\left(z_{1},z_{2}\right)=\left(a,\overline{a}\right)}$ can be written in terms of the Wirtinger derivative of $f$, as $\left(\Re\left.\dfrac{\partial f(z)}{\partial z}\right|_{z=a},\Im\left.\dfrac{\partial f(z)}{\partial z}\right|_{z=a}\right)$ and similarly for $\dfrac{\partial\widehat{\mathbf{f}}(z_{1},z_{2})}{\partial z_{2}}$ and $\dfrac{\partial g(z)}{\partial\overline{z}}$.
Chain Rule
Let $a$ be a complex number with real and imaginary parts $(x,y)$. For convenience, set $H:=D\left(\mathbf{h}\right)=\dfrac{1}{2}\begin{bmatrix}1 & 1\\-i & i\end{bmatrix}$.
We have
\begin{align*}
&\phantom{=}\begin{bmatrix}\Re\left.\dfrac{\partial\left(g\circ f\right)(z)}{\partial z}\right|_{a} & \Re\left.\dfrac{\partial\left(g\circ f\right)(z)}{\partial\overline{z}}\right|_{a}\\
\Im\left.\dfrac{\partial\left(g\circ f\right)(z)}{\partial z}\right|_{a} & \Im\left.\dfrac{\partial\left(g\circ f\right)(z)}{\partial\overline{z}}\right|_{a}
\end{bmatrix}\\
&=D\left(\widehat{\mathbf{g\circ f}}\right)_{\left(a,\overline{a}\right)}=D\left(\widetilde{\mathbf{g\circ f}}\circ\mathbf{h}\right)_{\left(a,\overline{a}\right)}\\&=D\left(\widetilde{\mathbf{g\circ f}}\right)_{(x,y)}D\left(\mathbf{h}\right)_{\left(a,\overline{a}\right)}=D\left(\widetilde{\mathbf{g\circ f}}\right)_{(x,y)}H\\&=D\left(\widetilde{\mathbf{g}}\circ\widetilde{\mathbf{f}}\right)_{(x,y)}H=D\left(\left(\widehat{\mathbf{g}}\circ\mathbf{h}^{-1}\right)\circ\left(\widehat{\mathbf{f}}\circ\mathbf{h}^{-1}\right)\right)_{(x,y)}H\\&=D\left(\widehat{\mathbf{g}}\circ\mathbf{h}^{-1}\right)_{\widehat{\mathbf{f}}(a,\overline{a})}D\left(\widehat{\mathbf{f}}\circ\mathbf{h}^{-1}\right)_{(x,y)}H\\&=\left(D\left(\widehat{\mathbf{g}}\right)_{\mathbf{h}^{-1}\left(\widehat{\mathbf{f}}(a,\overline{a})\right)}H^{-1}\right)\left(D\left(\widehat{\mathbf{f}}\right)_{\left(a,\overline{a}\right)}H^{-1}\right)H\\&=D\left(\widehat{\mathbf{g}}\right)_{\left(f(a),\overline{f(a)}\right)}\left(H^{-1}D\left(\widehat{\mathbf{f}}\right)_{\left(a,\overline{a}\right)}\right)\\&=\begin{bmatrix}\Re\left.\dfrac{\partial g(z)}{\partial z}\right|_{f(a)} & \Re\left.\dfrac{\partial g(z)}{\partial\overline{z}}\right|_{f(a)}\\
\Im\left.\dfrac{\partial g(z)}{\partial z}\right|_{f(a)} & \Im\left.\dfrac{\partial g(z)}{\partial\overline{z}}\right|_{f(a)}
\end{bmatrix}\left(\begin{bmatrix}1 & i\\
1 & -i
\end{bmatrix}\begin{bmatrix}\Re\left.\dfrac{\partial f(z)}{\partial z}\right|_{a} & \Re\left.\dfrac{\partial f(z)}{\partial\overline{z}}\right|_{a}\\
\Im\left.\dfrac{\partial f(z)}{\partial z}\right|_{a} & \Im\left.\dfrac{\partial f(z)}{\partial\overline{z}}\right|_{a}
\end{bmatrix}\right)\\&=\begin{bmatrix}\Re\left.\dfrac{\partial g(z)}{\partial z}\right|_{f(a)} & \Re\left.\dfrac{\partial g(z)}{\partial\overline{z}}\right|_{f(a)}\\
\Im\left.\dfrac{\partial g(z)}{\partial z}\right|_{f(a)} & \Im\left.\dfrac{\partial g(z)}{\partial\overline{z}}\right|_{f(a)}
\end{bmatrix}\begin{bmatrix}\left.\dfrac{\partial f(z)}{\partial z}\right|_{a} & \left.\dfrac{\partial f(z)}{\partial\overline{z}}\right|_{a}\\
\left.\dfrac{\partial\overline{f}(z)}{\partial z}\right|_{a} & \left.\dfrac{\partial\overline{f}(z)}{\partial\overline{z}}\right|_{a}
\end{bmatrix}\\&=\begin{bmatrix}\Re\left.\dfrac{\partial g(z)}{\partial z}\right|_{f(a)}\left.\dfrac{\partial f(z)}{\partial z}\right|_{a}+\Re\left.\dfrac{\partial g(z)}{\partial\overline{z}}\right|_{f(a)}\left.\dfrac{\partial\overline{f}(z)}{\partial z}\right|_{a} & \Re\left.\dfrac{\partial g(z)}{\partial z}\right|_{f(a)}\left.\dfrac{\partial f(z)}{\partial\overline{z}}\right|_{a}+\Re\left.\dfrac{\partial g(z)}{\partial\overline{z}}\right|_{f(a)}\left.\dfrac{\partial\overline{f}(z)}{\partial\overline{z}}\right|_{a}\\
\Im\left.\dfrac{\partial g(z)}{\partial z}\right|_{f(a)}\left.\dfrac{\partial f(z)}{\partial z}\right|_{a}+\Im\left.\dfrac{\partial g(z)}{\partial\overline{z}}\right|_{f(a)}\left.\dfrac{\partial\overline{f}(z)}{\partial z}\right|_{a} & \Im\left.\dfrac{\partial g(z)}{\partial z}\right|_{f(a)}\left.\dfrac{\partial f(z)}{\partial\overline{z}}\right|_{a}+\Im\left.\dfrac{\partial g(z)}{\partial\overline{z}}\right|_{f(a)}\left.\dfrac{\partial\overline{f}(z)}{\partial\overline{z}}\right|_{a}
\end{bmatrix}\end{align*}
This yields the following two equations: $$\left.\dfrac{\partial\left(g\circ f\right)(z)}{\partial z}\right|_{a}=\left.\dfrac{\partial g(z)}{\partial z}\right|_{f(a)}\left.\dfrac{\partial f(z)}{\partial z}\right|_{a}+\left.\dfrac{\partial g(z)}{\partial\overline{z}}\right|_{f(a)}\left.\dfrac{\partial\overline{f}(z)}{\partial z}\right|_{a}$$ $$\left.\dfrac{\partial\left(g\circ f\right)(z)}{\partial\overline{z}}\right|_{a}=\left.\dfrac{\partial g(z)}{\partial z}\right|_{f(a)}\left.\dfrac{\partial f(z)}{\partial\overline{z}}\right|_{a}+\left.\dfrac{\partial g(z)}{\partial\overline{z}}\right|_{f(a)}\left.\dfrac{\partial\overline{f}(z)}{\partial\overline{z}}\right|_{a}$$ Using shorthand, these can be rewritten in the more familiar form: $$\dfrac{\partial\left(g\circ f\right)}{\partial z}=\left(\dfrac{\partial g}{\partial z}\circ f\right)\dfrac{\partial f}{\partial z}+\left(\dfrac{\partial g}{\partial\overline{z}}\circ f\right)\dfrac{\partial\overline{f}}{\partial z}$$ $$\dfrac{\partial\left(g\circ f\right)}{\partial\overline{z}}=\left(\dfrac{\partial g}{\partial z}\circ f\right)\dfrac{\partial f}{\partial\overline{z}}+\left(\dfrac{\partial g}{\partial\overline{z}}\circ f\right)\dfrac{\partial\overline{f}}{\partial\overline{z}}$$
