Differentiability of a piecewise complex function Consider
$$f(z) :=\begin{cases}z^3/\overline{z}^2, &z\neq0\\0, &z=0\end{cases} $$
Must show it satisfies Cauchy-Riemann equations at $z=0$. I can write it as $z^5/|z|^4$, but extracting $u,v$ from $(x+iy)^5/(x^2+y^2)^2$ seems too tedious. Alternatively I could write it as $r\cos 5\varphi + ir\sin 5\varphi$, which looks more appealing, But, does that mean $u = r\cos 5\varphi$? Do I proceed by implicitly differentiating $u$ w.r.t $x$? (analogously for $v$)
The second part of the problem tells us to show $f$ is not differentiable at $z=0$ and whether this contradicts the theory. My conclusion is, if the equations are satisfied And both $u,v$ are differentiable at $(x,y)$ then $f$ is differentiable. Since contradiction of theory is not in question, we must show either of $u,v$ are not differentiable at $0$.
How does one do that, though? 

To check the Cauchy-Riemann equations we only need to compute the partial derivatives. For the given function, the partial derivatives at $0$ are easily computed,
$$\frac{\partial f}{\partial x}(0) = \lim_{x\to 0} \frac{f(x) - f(0)}{x} = \lim_{x\to 0} \frac{1}{x}\cdot f(x) = \lim_{x\to 0} \frac{1}{x}\cdot \frac{x^3}{x^2} = 1,$$
and $f(iy)$ is for $y\in \mathbb{R}$ similarly nice, so $\frac{\partial f}{\partial y}(0)$ is also easily computed. Splitting $f$ into its real and imaginary parts before computing the partial derivatives is more complicated.
Rather than looking at the real and imaginary parts in isolation, it is more convenient to show that $f$ is not real differentiable at $0$. From the above, we know that the only candidate for the real differential is $\frac{\partial f}{\partial z}(0) = 1$ and $\frac{\partial f}{\partial \overline{z}}(0) = 0$, which then easily shows the non-differentiability.
