Solution: 1. Note that if $z\neq 0$ then \[ f(z) = \frac{\bar{z} ^3}{z^2}. \] As it is quotient of two continuous functions, and denominator never vanishes, if $z\neq 0$, so it is continuous for all $z\neq 0$. Thus, we only need to prove that $f(z)$ is continuous at $z=0$. Observe that \begin{align*} \lim_{z \to 0} \left\vert \frac{\bar{z} ^3}{z^2} \right\vert & = \lim_{z \to 0} \frac{\left\vert \bar{z} ^3 \right\vert }{\left\vert z^2 \right\vert } \\ & = \lim_{z \to 0} \frac{\left\vert z^3 \right\vert }{\left\vert z^2 \right\vert } \\ & = \lim_{z \to 0} \vert z \vert = 0. \end{align*} Thus, we got \[ \lim_{z \to 0} \left\vert \frac{\bar{z} ^3}{z^2} \right\vert = 0 \implies \lim_{z \to 0} \frac{\bar{z} ^3}{z^2} = 0 = f(0). \] Hence, $f$ is continuous at $z=0$.

2. We want to prove that the function is not differentiable at $z=0$. Note that \[ f^\prime (z_0) = \lim_{z \to z_0} \frac{f(z)-f(z_0)}{z-z_0}. \] We will prove that the above limit will not exist at $z=0$. Consider \begin{align*} \lim_{z \to 0} \frac{f(z)-f(0)}{z} & = \lim_{z \to 0} \frac{\frac{\bar{z} ^3}{z^2}- 0 }{z} \\ & = \lim_{z \to 0} \frac{\bar{z} ^3}{z^3}. \end{align*} Let $z=x+\iota y$. Then if $y=0$, then $z\to 0$ is equivalent to $x\to 0$. So, \begin{align*} \lim_{x \to 0} \frac{x^3}{x^3} = 1. \end{align*} On the other hand, if $x=0$, then $y\to 0$ and we have \begin{align*} \lim_{y \to 0} \frac{(-\iota y)^3}{(\iota y) ^3} & = \lim_{y \to 0} \frac{-\iota ^3}{\iota ^3}=-1. \end{align*} Therefore, the limit $\displaystyle \lim_{z \to 0} \frac{\bar{z} ^3}{z^3}$ does not exist and hence $f^\prime (0)$ does not exist.