Let $f(z) =zRe(z)$. Determine all points $z_0$ for which the complex derivative $f'(z_0)$ existst.
I wrote $f(z)$ as $f(z)=f(x+iy)=(x+iy)Re(x+iy)=x^2+(xy)i:=u(x)+v(x,y)i$.
So we get the partials $u_x=2x$, $v_y = x$, $u_y=0$, $v_x=y$.
Now the CR equations, $u_x=v_y$ and $u_y=-v_x$ , only hold in the point $z=0$.
So $f(z)$ is only diffb. in the point $z=0$. So $f'$ exists only in the point $0$;.
Apparantly this reasoning is wrong. But I don't know what is wrong about it?
What you have done shows that $f$ is not differentiable at any point other than $0$. The validity of C-R equations at $0$ does not guarantee existence of $f'(0)$. So you have to check existence of $f'(0)$ from definition: $f'(0)=\lim_{z \to 0} \frac {z Re(z) -0} z=\lim_{z \to 0} Re (z)=0$. Hence $f'(0)=0$.
Note: The validity of C-R equation at all points of an open set implies differentiability but validity of C-R equation at a point does not imply differentiability at that point.