Show that $f$ is continuous at $0$ and it satisfies the Cauchy Riemman conditions but it is not differentiable.
The following routine calculation proves continuity at $z=0$, switching to polar coordinates at the end:
$\vert f(z)-0\vert ^{2}=\left | \frac{x^{3}-y^{3}+i(x^{3}+y^{3})}{x^2+y^2} \right |^{2}=\frac{(x^{3}-y^{3})^{2}+(x^{3}+y^{3})^{2}}{(x^{2}+y^{2})^{2}}=2\frac{x^{6}+y^{6}}{(x^{2}+y^{2})^{2}}= 2\cdot \frac{2r ^{6}(\cos^6t+\sin^6t)}{4r ^{4}}=r^2(\cos^{6}t+\sin^{6}t)\to 0\ \text{as}\ r\to 0.$
For non-differentiability at zero, we observe that along $y=0$ we have, whenever $x\neq 0$:
$\left | \frac{f(z)}{z} \right |^{2}=\frac{x^{3}(1+i)}{x^{3}}=\left ( 1+i \right )$
and along $x=0$ whenever $y\neq 0$:
$\left | \frac{f(z)}{z} \right |^{2}=\frac{y^{3}(-1+i)}{y^{3}}=\left ( -1+i \right )$
and therefore $\lim _{z\to 0}\frac{f(z)}{z}$ does not exist.