To check differentiability of $f(z)=z |z|$ at $(0,0).$ Here $f(z)=z |{z}|=(x+iy) \sqrt{x^2+y^2}$.Here CR equations are not satisfied.Then to check differentiability at $(0,0),$  I find the partial derivatives $u_x,u_y,v_x,v_y$ at $(0,0)$ and found that CR equations are  satisfied at $(0,0)$ (Here $u_x,u_y,v_x,v_y$ are all zero at $(0,0).$) Hence given function is differentiable only at $(0,0).$
Any improvement in my answer and is there any other way to solve this.

Let's use the definition:
$$
f'(0)=\lim_{z\to 0}\frac{f(z)-f(0)}{z}=\lim_{z\to 0}|z|=0
$$
So $f'(0)$ exists and is $0$.
We can argue by contradiction at $z≠0$. Indeed, let $z_0≠0$ and assume $f$ is differentiable at $z_0$. Then so is the function $g(z):=\frac{f(z)}{z}$. But $g(z)=|z|$, which is a contradiction since $z\mapsto |z|$ is nowhere differentiable.
