Prove that the function $f(z) = \sqrt{|Re(z) Im(z)|}$ satisfies the Cauchy-Riemann equations I'm asked the following question:

Prove that the function $f: \mathbb{C} \to \mathbb{C}$ defined by
$$f(z) = \sqrt{|Re(z) Im(z)|}$$ satisfies the Cauchy-Riemann equation,
  but not differentiable there.

However, the function is basically $f(x+iy) = \sqrt{|xy|},$ i.e $f(z) = u + iv$, $u = \sqrt{|xy|}$ and $v = 0$.
However, for example, the partial derivatives of $f$ wrt $x$ is $\frac{\pm \sqrt{y} }{ 2|x|} $, and this function does not have a limit at the origin, so $\frac{\partial u}{\partial x } $ does not exist at the origin, hence how can it satisfy the Cauchy-Riemann equation ?
However

When you calculate partial derivative of $f$ at $(0,0)$ you have to set $y=0$. So the partial derivative w.r.t. $x$ is $\lim _{x \to 0} \frac{0-0} x$ which is $0$.
