Cauchy-Riemann Equation for Surface Equations Suppose we are given smooth functions $W,V:\mathbb{R}^2\times\mathbb{C}\to\mathbb{C}^2$ and are asked to determine if the function $Q(W(x,y,z(x,y)),V(x,y,z(x,y))$ solving
$$
Q(W(x,y,z(x,y)),V(x,y,z(x,y)))=0,
$$
is analytic. Following Chapter 10 of Lectures on Random Lozenge  Tilings, Gorin states that the sufficient condition for $Q$ to be analytic is to establish $W(V)$ is analytic or in other words
$$\frac{\partial W}{\partial\bar{V}}=0.
$$
Of course this form of the Cauchy-Riemann equations
$$\frac{\partial f}{\partial \bar{z}}=0
$$
is recognisable when establishing $f(z):\mathbb{C}\to\mathbb{C}$ is analytic, but I don't understand how $\frac{\partial W}{\partial\bar{V}}=0$ implies $Q$ is analytic. What's the basic reason for this? Also, would $\frac{\partial V}{\partial\bar{W}}=0$ do the job, too?
For context, see the latter portion of Chapter 9 from the linked reference which derives the surface equation $Q=0$ from the method of complex characteristics and where there is a concrete example of the functions $W,V$.
 A: I think establishing function $f(z): \mathbb{C} \rightarrow \mathbb{C}$,  mentioned by you, is needed for our consideration to answer your question.
We have, that partial derivatives $u_x$,$u_y$,$v_x$,$v_y$ are continuous(by smoothness in your case), and we have, generally, that your function Q pass circles in circles,particularly, circle with center in $z_0 \in D$ into circle with center in $\overline{z_0} \in D$(we can consider your curves W and V from Gorin's book like bounded Jordan curves, and, within it we can split it on infinite amount of circles, from smoothness and Jordan lemmas).
$\omega - \omega_0 = f_{z_0}(z-z_0) + f_\overline{z_0}(\overline{z}-\overline{z_0})$ - main linear part of that mini-functions, where
$ f_{z_0} = \frac{\partial f}{\partial z}_{z=z_0}$ and
$f_{\overline{z_0}} = \frac{\partial f}{\partial \overline{z}}_{z=z_0}$
and, because of that, $f(z)$ is analytic or anti-analytic in $D$.
What's that got to do with your question? So, if you write down square of module of $\omega - \omega_0$, you will get
$
 \left\lvert  \omega  - \omega_0   \right\rvert^{2} = (\left\lvert f_{z_0}\right\rvert^{2} + \left\lvert f_{\overline{z_0}}\right\rvert^{2})\left\lvert z-z_0   \right\rvert^{2}+2Re [f_{z_0}\overline{f_{\overline{z_0}}}(z-z_0)^{2}]$
thus, circle $\left\lvert z-z_0 \right\rvert = r_1$ goes to $\left\lvert \omega-\omega_0 \right\rvert = r_2$, in last equation we will have
$f_{z_0}\overline{f_{\overline{z_0}}} = 0$, wherein $f_{z_0} = 0$,$f_{\overline{z_0}} \ne 0$ or $f_{z_0} \ne 0$,$f_{\overline{z_0}} = 0$.
