$f(z) = |z|^2$ is complex differentiable only on $(0,0)$ Here's my proof attempt:

$$f(z) = |z|^2$$

When $z=x+yi$, we have:

$$f(z) = x^2+y^2 + 0i$$

Doing Cauchy Riemann

$$\frac{\partial u}{\partial x} = 2x = \frac{\partial v}{\partial y} = 0$$

only when $x=0$.
The second Cauchy Riemann condition is

$$\frac{\partial v}{\partial x} = 0 = -\frac{\partial u}{\partial y} = 2y$$

only when $y = 0$.

Therefore, the function can be differentiable only on $(0,0)$ but we haven't proven it yet, we must show that the partial derivatives are continuous on $(0,0)$, but they are. Therefore, $f(z)$ is only complex differentiable on $(0,0)$

Is everything all right?
It looks good except when you say "we must show that the partial derivatives are continuous". It would be enough to  prove that $f$ is differentiable as a real function of two variables at the origin.
It looks good except when you say "we must show that the partial derivatives are continuous". It would be enough to  prove that $f$ is differentiable as a real function of two variables at the origin.