If $f(z)$ is differentialble at $z_0$ then the following limit \begin{eqnarray*} f'(z_0) & = & \lim_{\xi \to 0} \frac{f(z_0+\xi)-f(z_0)}{\xi} \end{eqnarray*}will remain the same approaching from any direction. First we fix $\theta$ as $\theta_0$ then we take the limit along the ray where the argument is equal to $\theta_0$ Then \begin{eqnarray*} f'(z_0) & = & \lim_{h \to 0} \frac{f(r_0e^{i\theta_0} + he^{i\theta_0})-f(r_0e^{i\theta_0})}{he^{i\theta_0}} \\ & = & \lim_{h\to 0} \frac{f((r_0+h)e^{i\theta_0})-f(r_0e^{i\theta_0})}{he^{i\theta_0}} \\ & = & \lim_{h\to 0} \frac{u(r_0+h,\theta_0) + iv(r_0+h,\theta_0) - u(r_0,\theta_0) - iv(r_0,\theta_0)}{he^{i\theta_0}}\\ & = & \frac{1}{e^{i\theta_0}} \Bigg[ \lim_{h\to 0} \frac{u(r_0+h,\theta_0) - u(r_0,\theta_0)}{h} + i \lim_{h\to 0} \frac{v(r_0+h,\theta_0) - v(r_0,\theta_0)}{h} \Bigg]\\ & = & \frac{1}{e^{i\theta_0}}\Bigg[ \frac{\partial u}{\partial r}(r_0,\theta_0) + i\frac{\partial v}{\partial r}(r_0,\theta_0) \Bigg] \end{eqnarray*} Similarly, if we take the limit along the circle with fixed $r$ equals $r_0$ Then

\begin{eqnarray*} f'(z_0) & = & \lim_{h\to 0} \frac{f(r_0e^{i\theta_0} + r_0e^{i(\theta_0+h)})-f(r_0e^{i\theta_0})}{r_0e^{i\theta_0}(e^{ih}-1)}\\ & = & \lim_{h\to 0} \frac{f(r_0e^{i(\theta_0+h)})-f(r_0e^{i\theta_0})}{he^{i\theta_0}}\\ & = & \lim_{h\to 0} \frac{u(r_0,\theta_0+h) + iv(r_0,\theta_0+h) -u(r_0,\theta_0) - iv(r_0,\theta_0)}{he^{i\theta_0}}\\ & = & \frac{1}{r_0e^{i\theta_0}} \Bigg[ \lim_{h\to 0} \frac{u(r_0+h,\theta_0) - u(r_0,\theta_0)}{h}\cdot \frac{h}{e^{ih}-1} + i \lim_{h\to 0} \frac{v(r_0+h,\theta_0) - v(r_0,\theta_0)}{h}\cdot \frac{h}{e^{ih}-1} \Bigg]\\ & = & \frac{1}{r_0e^{i\theta_0}} \Bigg[ \lim_{h\to 0} \frac{u(r_0+h,\theta_0) - u(r_0,\theta_0)}{h}\cdot \lim_{h\to 0} \frac{h}{e^{ih}-1} + i \lim_{h\to 0} \frac{v(r_0+h,\theta_0) - v(r_0,\theta_0)}{h}\cdot \lim_{h\to 0} \frac{h}{e^{ih}-1} \Bigg]\\ & = & \frac{1}{r_0e^{i\theta_0}} \Bigg[ \frac{\partial u}{\partial \theta}(r_0,\theta_0)\frac{1}{i} + \frac{\partial v}{\partial \theta}(r_0,\theta_0) \Bigg]\\ & = & \frac{1}{r_0e^{i\theta_0}} \Bigg[ \frac{\partial v}{\partial \theta}(r_0,\theta_0) - i\frac{\partial u}{\partial \theta}(r_0,\theta_0) \Bigg] \end{eqnarray*} Note: We use l'Hôpital's rule to obtain the following result used above $\lim_{h\to 0} \frac{h}{e^{ih}-1} = \frac{1}{i}$

Now, since the limit is the same along the circle and the ray then they are equal: \begin{eqnarray*} \frac{1}{e^{i\theta_0}}\Bigg[ \frac{\partial u}{\partial r}(r_0,\theta_0) + i\frac{\partial v}{\partial r}(r_0,\theta_0) \Bigg] & = & \frac{1}{r_0e^{i\theta_0}} \Bigg[ \frac{\partial v}{\partial \theta}(r_0,\theta_0) - i\frac{\partial u}{\partial \theta}(r_0,\theta_0) \Bigg]\\ \Bigg[ \frac{\partial u}{\partial r}(r_0,\theta_0) + i\frac{\partial v}{\partial r}(r_0,\theta_0) \Bigg] & = & \frac{1}{r_0} \Bigg[ \frac{\partial v}{\partial \theta}(r_0,\theta_0) - i\frac{\partial u}{\partial \theta}(r_0,\theta_0) \Bigg]\\ \end{eqnarray*}which implies that \begin{eqnarray*} \frac{\partial u}{\partial r} & = & \frac{1}{r}\frac{\partial v}{\partial \theta}\\ \frac{\partial v}{\partial r} & = & -\frac{1}{r}\frac{\partial u}{\partial \theta} \end{eqnarray*}QED