Prove that $f(x)$ satisfies the Cauchy-Riemann Equations at the point $z=0$
$$f(z) = \begin{cases} \dfrac{\overline{z}^3}{|z|^2}, & \text{if $z$ is not 0} \\ 0, & \text{if $z$ = 0} \end{cases} $$
For $z=x+iy$ i get $u=\frac {x^3-3xy^2}{x^2+y^2}, v=\frac {y^3-3yx^2}{x^2+y^2} $ Which do not satisfy the C-R Equations. I also tried with polar form which also didn't work. I supect it has to do with the fact that i have to chech the C-R equations on the point $z=0+i0$ but since the function is $0$ at that point, i don't know what to do.
Since $u(x,y)=\dfrac{x^3-3xy^2}{x^2+y^2}$, you have $u(x,0)=x$ and $u(0,y)=0$. Therefore, $\dfrac{\partial u}{\partial x}u(0,0)=1$ and $\dfrac{\partial u}{\partial y}(0,0)=0$. On the other hand, since $v(x,y)=\dfrac{y^3-3x^2y}{x^2+y^2}$, you have $v(x,0)=0$ and $v(0,y)=y$. Therefore, $\dfrac{\partial v}{\partial x}u(0,0)=0$ and $\dfrac{\partial v}{\partial y}(0,0)=1$. So, yes, $(0,0)$ is a solution of the Cauchy-Riemann equations.