Solution: The given function is \[ f(z) = \begin{cases} \frac{x^2 - y^2 - 2xy \iota }{x + \iota y}, &\text{ if } z \neq 0 ;\\ 0, &\text{ if } \text{ otherwise}. \end{cases} \] Note that $\bar{z} ^2 = (x - \iota y)^2 = x^2 - 2xy \iota - y^2$. Thus, \begin{align*} f(z) = \begin{cases} \frac{\bar{z} ^2}{z}, &\text{ if } z\neq 0 ;\\ 0, &\text{ if } \text{ otherwise}. \end{cases} \end{align*} On of the formulation of the Cauchy-Riemann equations is $\frac{\partial f}{\partial \bar{z} } = 0 $. Note that \begin{align*} \frac{\partial f}{\partial \bar{z} } = \frac{2 \bar{z} }{z}. \end{align*} Since $\displaystyle \lim_{z \to 0} \frac{\partial f}{\partial \bar{z} } $ does not exist, and hence the function does not satisfy the Cauchy-Riemann equation at $z = 0$.

We can also solve this by writing $f = u + \iota v$. \begin{align*} f(z) & = \frac{\bar{z} ^2}{z} \times \frac{\bar{z} }{\bar{z} } = \frac{\bar{z} ^3}{\vert z \vert ^2} \\ & = \frac{(x - \iota y)^3}{x^2 + y^2} \\ & = \frac{x^3 + 3x y^2}{x^2 + y^2} + \iota \left( \frac{y^3 - 3x^2 y}{x^2 + y^2} \right). \end{align*} Thus, \[ u(x,y) = \frac{x^3 + 3x y^2}{x^2 + y^2}\quad \text{ and } \quad v(x,y) = \frac{y^3 - 3x^2 y}{x^2 + y^2}. \] Recall that the Cauchy-Riemann equations at $z_0 = (x_0, y_0)$ are \[ u_x(z_0) = v_y(z_0) \text{ and } u_y(z_0) = - v_x(z_0). \] We need to check the Cauchy-Riemann equations at $z = 0$. Note that \begin{align*} u_x(0,0) & = \lim_{h \to 0} \frac{u((h,0)) - u(0,0)}{h} = \frac{h}{h} = 1\\ v_x(0,0) & = \lim_{h \to 0} \frac{v((h,0)) - v(0,0)}{h} = 0\\ u_y(0,0) & = \lim_{k \to 0} \frac{u((0,k)) - u(0,0)}{k} = 0 \\ v_y(0,0) & = \lim_{k \to 0} \frac{v((0,k)) - v(0,0)}{k} = \frac{h}{h} = 1. \end{align*} Thus, $f$ satisfies the Cauchy-Riemann equations at $z = (0,0)$.

We now will check the differentiability of $f$ at $(0,0)$. Recall that $f$ is differentiable at $z = 0$ if the following limit exists \begin{align*} \lim_{z \to 0} \frac{f(z) - f(0)}{z} & = \lim_{z \to 0} \frac{\frac{\bar{z} ^2}{z} - 0}{z} \\ & \lim_{z \to 0} \frac{\bar{z} ^2}{z^2}. \end{align*} We will evaluate this limit in two different directions. Let $h \in \mathbb{R} $. \begin{align*} \lim_{h \to 0} \frac{\left( \overline{h + \iota 0} \right) ^2}{(h + \iota 0)^2} = 0. \end{align*} On the other hand, \begin{align*} \lim_{h \to 0} \frac{\left( \overline{h + \iota h} \right) ^2}{(h + \iota h)^2} & = \lim_{h \to 0} \frac{h^2 (1 - \iota )^3}{h^2 ( 1 + \iota )^2} \\ & = \frac{(1 - \iota )^2}{(1 + \iota )^2} = -1. \end{align*} Thus, $f$ is not differentiable at $z = 0$.