Square root complex function $f(x+iy)=\sqrt{|x||y|} $ for all $x,y \,\,\epsilon \,\, \mathbb R$. 
Show that $f$ satisfies Cauchy-Riemann equations at $0$ but is not analytic at $0$.
Attempt at a solution:
Since the function has only a real part, $Re\,f= u(x,y)=\sqrt{|x||y|}$ and $Im\,f=v(x,y)=0.$ Need to show that all partial derivatives are $0$ and discontinuous at $0$. But taking the partial derivatives gives $\frac{\partial u}{\partial x}=\frac{1}{2}\frac{\sqrt{|y|}}{\sqrt{|x|}}$ which does not exist at $0$.
Any help is much appreciated. 

$\frac{\partial u}{\partial x}(0,0)=\lim_{t\to0}\frac{u(0+t,0)-u(0,0)}{t}=\lim_{t\to0}0=0$
$\frac{\partial u}{\partial y}(0,0)=\lim_{t\to0}\frac{u(0,0+t)-u(0,0)}{t}=\lim_{t\to0}0=0$
If $x>0$ and $y>0$ then:
$\frac{\partial u}{\partial x}(x,y)=\frac{y}{2\sqrt{xy}}=\frac{1}{2}\sqrt{\frac{y}{x}}$. If we let $y=\alpha x$ in this region then:
$\frac{\partial u}{\partial x}(x,\alpha x)=\frac{1}{2}\sqrt{\frac{\alpha x}{x}}=\frac{\sqrt{\alpha}}{2}$.
This means that if we approach $(0,0)$ through the line $y=\alpha x$ then the partial derivatives in $x$ at these points approaches $\frac{\sqrt{\alpha}}{2}$. Hence the partial derivatives in $x$ are not continuous at $0$. A similar thing happens for the partials in $y$.
