$\sqrt{|xy|}$ verifies the Cauchy-Riemann equations at $(0,0)$, but is not differentiable at $(0,0)$ 
Show that the function $f(z)=\sqrt{|xy|}$ verifies the Cauchy-Riemann equations at point zero, but it is not differentiable at that point.

$u(x,y)=\sqrt{|xy|}$ and $v(x,y)=0$ since the imaginary part of the function is zero.
$\frac{\partial u(x,y)}{\partial x}=\frac{1}{2}\sqrt{\frac{y}{x}}\\\frac{\partial v(x,y)}{\partial y}=0$
$\frac{\partial u(x,y)}{\partial y}=\frac{1}{2}\sqrt{\frac{x}{y}}\\\frac{\partial v(x,y)}{\partial x}=0$
$\begin{cases}
\frac{1}{2}\sqrt{\frac{y}{x}}=0\\
\frac{1}{2}\sqrt{\frac{x}{y}}=0\end{cases},$
Its solution should be (0,0) but $\frac{0}{0}$ is not determined.
Then to check it is not differentiable at point (0,0) I used the derivative definition:
$\lim_{z\to 0}\frac{f(z)-f(0)}{z-0}=\frac{\sqrt{|xy|-0}}{x-0+i(y-0)}=\frac{0}{0}$ again not determined.
Questions:
Is my solution right? According to the question the function should verify the Cauchy-Riemann equations however it does not seem so. What am I doing wrong on that part?
Thanks in advance!
 A: No, your solution isn't right. You are supposed to compute $\frac{\partial u}{\partial x}(0,0)$ and $\frac{\partial u}{\partial y}(0,0)$, which you did not do. Just note that$$\frac{\partial u}{\partial x}(0,0)=\lim_{h\to0}\frac{u(h,0)-u(0,0)}h=\lim_{h\to0}\frac{0-0}h=0$$amd, for the same reason, $\frac{\partial u}{\partial y}(0,0)=0$. So, yes, $(0,0)$ is a solution of the Cauchy-Riemann equations.
This proves that, if $f$ is differentiable at $0$, then $f'(0)$ can only be equal to $0$. However, of $z$ is of the form $x+xi$ (with $x\in(0,\infty)$), then$$\frac{f(x+xi)}{x+xi}=\frac x{x+xi}=\frac1{1+i},$$and therefore we can't have $f'(0)=0$. So, a contradiction is reached, which proves that $f$ is not differentiable at $0$.
A: No, your solution is not correct. To verify that C-R equations hold at $(0,0)$ you have to calculate the partial derivative at this point from definition. For example, $\frac {\partial } {\partial  x} u(0,0)=\lim \frac {u(h,0)-u(0,0)} h$ which is $0$. Simailarly all partial derivatives at the origin are $0$ so C-R equations are satisfied.
To show that $f$ is not differentiable at $(0,0)$  note that $\frac {f(\frac 1 n +i\frac 1 n )-f(0,0)} {\frac 1 n +i\frac 1 n } \to \frac 1 {1+i}$ whereas $[f(x,0)-f(0,0)]/x \to 0$ as $x \to 0$. 
A: At $(x,y)=(0,0)$
$$\dfrac{\partial u(x,y)}{\partial x}=\lim_{x\to 0}\dfrac{u(x,0)-u(0,0)}{x-0}=0$$
