Solution: Let us write the function \[ f(z) = \sqrt{\vert xy \vert } = u(x,y) + \iota v(x,y). \] So, $u(x,y) = \sqrt{\vert xy \vert } $ and $v(x,y)=0$. Note that \begin{align*} \left.\frac{\partial u}{\partial x}\right|_{(x,y)=(0,0)} & = \lim_{h \to 0} \frac{u(h,0)-u(0,0)}{h} \\ & = \lim_{h \to 0} \frac{0}{h} = 0. \end{align*}

Similarly, \begin{align*} \left.\frac{\partial u}{\partial y}\right|_{(x,y)=(0,0)} & = \lim_{k \to 0} \frac{u(0,k)-u(0,0)}{k} \\ & = \lim_{k \to 0} \frac{0}{k} = 0. \end{align*} Thus, \[ \frac{\partial u}{\partial x} = 0 = \frac{\partial v}{\partial y},~\text{ and } \frac{\partial u}{\partial y} = 0 = - \frac{\partial v}{\partial x}. \] Therefore, $f$ satisfies the Cauchy-Riemann equations at $z=0$.

Now we need ot show that $f$ is not differentiable at $z=0$. Consider the following limit \begin{align*} \lim_{h \to 0} \frac{\vert f(h,0)-f(0,0) \vert }{\vert (h,0) \vert } & = 0 = \lim_{k \to 0} \frac{\vert f(0,k)-f(0,0) \vert }{\vert (0,k) \vert }. \end{align*} If $f$ is differentiable at $z=0$, then the derivative would be zero (since if $f$ is differentiable, the derivative would be the partial derivatives in a matrix) but note that \begin{align*} \lim_{h \to 0} \frac{\vert f(h,h)-f(0,0)-0 \vert }{\vert (h,h) \vert } & = \lim_{h \to 0} \frac{\sqrt{h^2} }{\sqrt{2h^2} } = \frac{1}{\sqrt{2}} \end{align*} which is not equal to zero and hence $f$ is not differentiable at $z=0$.

Another way is to directly show that limit \[ \lim_{z \to 0} \frac{f(z)-f(0)}{z} \] does not exist. We will show that the above limit does not exist along the line $y=x$. \[ \lim_{x \to 0} \frac{f(x+\iota x)-0}{x+\iota x} = \lim_{x \to 0} \frac{\sqrt{\vert x\cdot x \vert } }{x+\iota x} = \left( \frac{1}{1+\iota } \right) \lim_{x \to 0} \frac{\vert x \vert }{x}, \] which does not exist. Thus, we proved that the function is not differentiable at $z=0$.