Formally we have the chain rule: $$(f\circ g)'(x)= (f'\circ g) \cdot g'$$
Let $f:\mathbb R^{3} \to \mathbb C$ and $g:\mathbb C\to \mathbb C$ be nice functions. Then does it make sense to talk that $$\frac{\partial}{\partial {x_1}} (g\circ f)(x)= (g'\circ f(x)) \cdot \frac{\partial}{\partial x_1} f(x), (*)$$ where $x=(x_1, x_2, x_3)$ and $g'(z)=\frac{d}{dz} g(z), z\in \mathbb C$ (complex derivative)
Note that we are mixing complex derivative first and at the end partial derivative.
My Question: (1)Is the above formula true? Does it make to sense to talk of the above $(*)$ formula? And what should be the multiplication $\cdot $ in the formula ? (2) Suppose now we consider $g$ as function of $z$ and $\bar{z}$ (eg. $g(z)=z^{2} \bar{z}^3$). Does it make sense to talk that $\frac{\partial}{\partial {x_1}} (g\circ f)(x)= (g'\circ f(x)) \cdot \frac{\partial}{\partial x_1} f(x),$ where $x=(x_1, x_2, x_3)$ and wghat should be $\cdot $ ?, and $g'(z)=\frac{\partial}{\partial z} g(z), z\in \mathbb C$ (complex derivative)
Yes, that is perfectly ok if by $g$ 'nice' you mean complex differentiable. If we restrict to the derivative in the direction $e_1$ then for $h$ (real and) small: $$ f(x+he_1)=f(x)+\partial_1f(x)h +o(h)$$ Since also (complex derivative) $$ g(z+\delta z) = g(z) + g'(z) \delta z+ o(\delta z)$$ we get (for the last you should do some 'little-o' estimate): $$g(f(x+he_1))= g(f(x) + \partial_1 f(x) h + o(h)) = g(f(x)) + g'(f(x)) \partial_1 f(x) h + o(h)$$ from which you can read off the precise formula (*) that you have stated.
If $g$ is a function of both $z$ and $\bar{z}$ then you get two terms, one from each partial derivative of $g$ + some complex conjugation. One has $$ g(z+\delta z) = g(z) + \partial_z g(z) \delta z+ \partial_{\bar{z}} g(z) \overline{\delta z} + o(\delta z)$$ and the derivative of the composed function becomes: $$ \frac{\partial}{\partial x_1} (g(f(x))) = \partial_z g(f(x)) \partial_1 f(x) + \partial_{\bar{z}} g(f(x)) \overline{\partial_1 f(x)}$$
For example, with $g(z)=z\; \bar{z}=|z|^2$ we have $$|z+\delta z|^2-|z|^2 = z \; \overline{\delta z} + \overline{z} \;\delta z \;+ o(\delta z)$$