I was working on a problem in field theory (so physics) and the question occured to me, wether the expression $\frac{\partial f^{*}}{\partial f}$ can be determined for a function $f:\mathbb{R}^n \rightarrow \mathbb{C} $ (generally).
I tried an approach inspired by Wirtinger derivatives (with $x,y: \mathbb{R}^{n} \rightarrow \mathbb{C}$):
$$ \textrm{using} \quad f = x + iy, \enspace f^{*} = x - iy \quad \textrm{and} \quad x= \frac{1}{2}(f+f^{*}), \enspace y=\frac{i}{2}(f^{*}-f) \textrm{:} $$
$$\frac{\partial f^{*}(x,y)}{\partial f} = \frac{\partial f^{*}}{\partial x} \frac{\partial x}{\partial f}+\frac{\partial f^{*}}{\partial y} \frac{\partial y}{\partial f} = 1 \cdot \frac{1}{2} + (-)i \cdot (-)\frac{i}{2} = 0 $$
Even though I'd like to believe thats true I am very much not sure (frankly I thing it looks wrong). Yet I can't see where I am mistaken (my knowledge of complex analysis is...let's say: bouned). So please let me know where I went wrong.
You have to go a step deeper and first define what the derivative of $f$ in the direction of $f^*$ could represent in terms of real partial derivatives and real functions $g,h$ of two real variables $x,y$. Terminology is
$$f(z) = g(x,y) + i h(x,y)$$ $$ f^*(z) = g(x, y) - i h(x, y)$$ $$ (f(z))^* = g(x ,- y) - i h(x ,- y)$$
As power series, this means that g and h have real coefficients
The fundamental Riemann equations for an analytic function in the space of general functions of two independent complex variables is
$$ 0 = \partial_{z^*} f(z,z*) = \partial_x g(x,y) - i \partial_y g(x,y) + i (\partial_x g(x,y) - i \partial_y g(x,y)) = 0$$
by linearity $$\partial_{a x + b y} f = a \partial_x f + b \partial_y f $$
That is f is diffentiable and independent of $x-iy$.
These are the concepts you are using as a primitive example, when x and y are the primary real variables in the complex plane.
As a first step, explain your idea for simplest nonlinear map.
$$\partial_{x^2-y^2 -2 i x y } (x^2 -y^2 + 2 i x y)$$
as a directional derivative when what is kept constant?