In $C\setminus \{0\}$,
Since $\frac{\partial u}{\partial x} = \frac{2x^{2}+y^{2}}{\sqrt{ x^{2}+y^{2}}}$ and $ \frac{\partial u}{\partial y}=\frac{xy}{\sqrt{ x^{2}+y^{2}}}$, $\frac{\partial v}{\partial x} = \frac{xy}{\sqrt{ x^{2}+y^{2}}}, \frac{\partial v}{\partial y} = \frac{x^{2}+2y^{2}}{\sqrt{ x^{2}+y^{2}}} $,
it can be easily observed that $f(z)=z|z|$ does not satisfy Cauchy-Riemann equations. I showed that $f(z)$ is differentiable at $z=0$, but I wonder whether we cannot discuss Cauchy-Riemann equations at $z=0$.
In this case, we have $u(x,y)=x\sqrt{x^2+y^2}$ and $v(x,y)=y\sqrt{x^2+y^2}$. So,\begin{align}\frac{\partial u}{\partial x}(0,0)&=\lim_{h\to0}\frac{h|h|}h=0\\\frac{\partial u}{\partial y}(0,0)&=\lim_{h\to0}0=0\\\frac{\partial v}{\partial x}(0,0)&=\lim_{h\to0}0=0\\\frac{\partial v}{\partial y}(0,0)&=\lim_{h\to0}\frac{h|h|}h=0.\end{align}So, $(0,0)$ is a solution of the Cauchy-Riemann equations. In general this is not enough to deduce that $f$ is differentiable at $0$, but it follows from your computations that $\frac{\partial u}{\partial x}$, $\frac{\partial u}{\partial y}$, $\frac{\partial v}{\partial x}$, and $\frac{\partial v}{\partial y}$ are continuous at $(0,0)$, and now we can deduce that $f$ is indeed differentiable at $0$. But it is much simpler to reach that conclusion without using the Cauchy-Riemann equations.