Prove that $f(z) = |z|^2$ is differentiable only at the origin

Another way of proving that $f$ is differentiable at $0$ is simply to observe that$$\lim_{z\to0}\frac{|z|^2}z=\lim_{z\to0}\overline z=0.$$Besides, if $z_0\neq0$, then$$\lim_{z\to z_0}\frac{|z|^2-|z_0|^2}{z-z_0}=\lim_{z\to z_0}\frac{|z|-|z_0|}{z-z_0}\bigl(|z|+|z_0|\bigr).$$Now, if $z$ approaches $z_0$ along the circle centered at $0$ passing through $z_0$, then the previous limit is $0$. And if $z$ approaches $z_0$ along the ray $\bigl\{\lambda z_0\,|\,\lambda\in(1,+\infty)\bigr\}$, then the previous limit is $2\overline{z_0}\neq0$. Therefore the limit does not exist.