The following problem is from the book A Collection of Problems on Complex Analysis by Volkovyskii, Lunts and Aramanovich. (The answer is omitted)
$\mathbf{110.}\,\,$ Prove that the Cauchy-Riemann equations are satisfied for the function $f(z)=\sqrt{|xy|}$ at the point $z=0$ but the derivative does not exist.
I started with the C-R equations being easier to prove:
$$\begin{aligned} & v_x=0.\\ &v_y=0.\\ &u_x=\lim_\limits{ \Delta x\to0}\frac{\sqrt{|(x+ \Delta x)y|}-\sqrt{|xy|}}{\Delta x}= \sqrt{|y|}\cdot\lim_\limits{ \Delta x\to0}\frac{\sqrt{|(x+ \Delta x)|}-\sqrt{|x|}}{\Delta x}\Big|_{(0,0)}=0\cdot\lim(\cdots)=0.\\ &u_y=\lim_\limits{ \Delta y\to0}\frac{\sqrt{|x(y+\Delta y)|}-\sqrt{|xy|}}{\Delta y}= \sqrt{|x|}\cdot\lim_\limits{ \Delta y\to0}\frac{\sqrt{|(y+ \Delta y)|}-\sqrt{|y|}}{\Delta y}\Big|_{(0,0)}=0\cdot\lim(\cdots)=0. \end{aligned}$$ Thus the Cauchy Riemann equations are satisfied at $z=0$. For the differentiability I am not sure if my work is correct.
We know that $$f’(z_0)=\lim_\limits{\Delta z\to 0}\frac{f(z_0+\Delta z)-f(z_0)}{\Delta z}$$
I was able to express the function in terms of $z$: $$f(z)=\sqrt{|xy|}=\frac12\sqrt{\left|z^2-\bar{z}^2\right|}$$ And applying the above definition $$f’(0)=\lim_\limits{\Delta z\to 0}\frac{\frac12\sqrt{\left|\Delta z^2-\Delta\bar{z}^2\right|}-0}{\Delta z} =\frac12 \lim_\limits{\Delta z\to 0}\frac{\sqrt{\left|\Delta z^2-\Delta\bar{z}^2\right|}}{\Delta z}= \frac{\sqrt2}2 \lim_\limits{(\Delta x,\Delta y)\to (0,0)}\frac{\sqrt{\left|\Delta x^2-\Delta y^2\right|}}{\Delta x + i\Delta y}$$ I chose 2 paths to evaluate the limits at: $1.\, \Delta y=\Delta x$ with $\Delta y\neq0$ and $2.\, \Delta y=0, \Delta x\neq0$.
Thus $$\begin{aligned} &1.\,\,\, \frac{\sqrt2}2 \lim_\limits{\Delta x\to 0}\frac{\sqrt{\left|\Delta x^2-\Delta x^2\right|}}{(1+i)\Delta x}=0.\\ &2.\,\,\, \frac{\sqrt2}2 \lim_\limits{\Delta x\to 0}\frac{\sqrt{\left|\Delta x^2\right|}}{\Delta x}= \frac{\sqrt2}2 \lim_\limits{\Delta x\to 0}\frac{\left|\Delta x\right|}{\Delta x}=\text{Does not exist}. \end{aligned}$$ Therefore we can conclude that the function is not differentiable at $z=0$ even though the Cauchy-Riemann equations hold at that point.
Could have I just evaluated $2$ and from there concluded that it is not differentiable at $z=0$? Is there a quicker way to see if it not differentiable at such point? Is there a way to compute $f’(z)$ with the limit definition stated earlier?