Establish the following generalization of the Cauchy-Riemann equations: if $f(z)=u(z)+iv(z)$ is differentiable at a point $z_0=x_0+iy_0$ on a domain $G\subset \mathbb{C}$, with $u,v:\mathbb{C}\to\mathbb{R}$, $f:\mathbb{C}\to\mathbb{C}$, then \begin{equation} \frac{\partial u}{\partial s}=\frac{\partial v}{\partial n}, \hspace{1cm}\frac{\partial u}{\partial n}=-\frac{\partial v}{\partial s} \end{equation} at $(x_0,y_0)$ where $\partial/\partial s$, $\partial/\partial n$ denote directional differentiation in any two orthogonal directions $s$ and $n$ such that $n$ is obtained from $s$ by making a counterclockwise rotation.
Proof for the Cauchy-Riemann equations: if $f(z)$ is differentiable at $z_0$, then \begin{equation}\Delta f(z)=f'(z_0)\Delta z + \epsilon\Delta z\end{equation} where $\epsilon\to 0$ as $\Delta\to 0$. Writing \begin{equation}f'(z_0)=a+ib, \hspace{1cm}\epsilon=\epsilon_1+\epsilon_2\end{equation} and taking the real and imaginary parts of the $\Delta$ equation above, we find that \begin{equation} \Delta u=a\Delta x-b\Delta y+\epsilon_1 \Delta_x - \epsilon_2\Delta y,\\ \Delta v=b\Delta x+a\Delta y +\epsilon_2 \Delta x + \epsilon_1 \Delta y\end{equation} where $\epsilon_1,\epsilon_2\to 0$ as $\Delta x, \Delta y\to 0$, since \begin{equation}|\Delta z|=\sqrt{(\Delta x)^2+(\Delta y)^2}, \hspace{.5cm} |\epsilon_1|\leq|\epsilon|,\hspace{0.3cm}|\epsilon_2|\leq|\epsilon|.\end{equation} It follows that dhe functions $u(x,y)$ and $v(x,y)$ are differentiable at $x_0+iy_0$ and so \begin{equation} \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \hspace{1cm}\frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} \end{equation} My initial trial for the proof: The Cauchy-Riemann equations are usually expressed in the standard basis of the complex plane, $e_1=1, e_2=i$, using \begin{equation}\Delta z=\Delta x+i\Delta y=[e_1\hspace{.3cm}e_2].[\Delta x, \Delta y]^T,\end{equation} however the equations of the question are expressed in the basis $[s\hspace{.3cm}n]$. By the definition of $s$ and $n$, if $s=s_a+is_b$ in the standard basis, then $n=-s_b+is_a$, and the matrix below provides a transformation from the standard basis to the basis $[s\hspace{.3cm}n]$: \begin{equation}\left[\begin{array}{cc}s_a & s_b\\-s_b & s_a\end{array}\right],\end{equation} which means that $\Delta z=(s_a\Delta x+s_b\Delta y)s+(-s_b\Delta x+s_a\Delta y)n$. From here, I got lost. I thought of substituting this new $\Delta z$ expression in the expression of $\Delta f(z_0)$ and then finding new expressions for $u$ and $v$ in terms of $s$ and $n$, but I could get nowhere on this line.
