Solution: If $z = x + \iota y$, then we note that \[ f(z) = u+ \iota v = \begin{cases} \frac{(x + \iota y)^5}{(x^2 + y^2)^2}, &\text{ if } z = 0;\\ 0, &\text{ if } z\neq 0. \end{cases} \] if $z \neq 0$. Then we have \begin{align*} \frac{\partial u}{\partial x} (0,0) & = \lim_{h \to 0} \frac{u(h,0) - u(0,0)}{h} \\ & = \lim_{h \to 0} \frac{h^5}{h^4 \times h} = 1. \\[2ex] \frac{\partial u}{\partial y} (0,0) & = \lim_{k \to 0} \frac{u(0,k) - u(0,0)}{k} \\ & = \lim_{k \to 0} \frac{0}{k^4 \times k } = 0. \\[2ex] \frac{\partial v}{\partial x} (0,0) & = \lim_{h \to 0} \frac{v(h,0) - v(0,0)}{h} \\ & = \lim_{h \to 0} \frac{0}{h^4 \times h} = 0. \\[2ex] \frac{\partial v}{\partial y} (0,0) & = \lim_{k \to 0} \frac{v(0,k) - v(0,0)}{k} \\ & = \lim_{k \to 0} \frac{k^5}{k^4 \times k } = 1. \end{align*} Therefore, \[ \frac{\partial u}{\partial x} (0,0) = 1 = \frac{\partial v}{\partial y} \text{ and } \frac{\partial u}{\partial y} = 0 = - \frac{\partial v}{\partial x}. \] Hence, $f$ satisfies the Cauchy-Riemann equations.

Now we will prove that $f$ is not differentiable at $z = 0$. Take $z = re^{\iota \theta}$ with $\theta = \frac{\pi}{4}$ and $r > 0$. Then \begin{align*} \lim_{z \to 0} \frac{f(z) - f(0)}{z} & = \lim_{r \to 0} \dfrac{\frac{r^5 e^{5\iota \theta }}{r^4}}{r e^{\iota \theta }} \\ & = \lim_{r \to 0} e^{4\iota \theta } = e^{\pi \iota } = -1. \end{align*} On the other hand, if $f$ is differentiable at $z=0$ , then \[ f'(0) = \frac{\partial u}{\partial x} (0) + \iota \frac{\partial v}{\partial x} (0) = 1. \] Thus, $f$ is not differentiable at $z = 0$.