Solution: We have \begin{align*} g(z) = \frac{1}{z} & = \frac{1}{x + \iota y} \\ & = \frac{1}{x -\iota y} \times \frac{x + \iota y}{x - \iota y} \\ & = \frac{x - \iota y}{x^2 + y^2} \\ & = \frac{x}{x^2 + y^2} + \iota \frac{-y}{x^2 + y^2}. \end{align*} Therefore, the real part of $g$ is $\frac{x}{x^2 + y^2}$ and imaginary part of $g$ is $\frac{-y}{x^2 + y^2}$.

We now will prove that the real part of $g$ satisfies the Laplace's equation. Let \begin{align*} u(x,y) = \frac{x}{x^2 + y^2} \implies \frac{\partial u}{\partial x} & = \frac{(x^2 + y^2)1 - x(2x)}{(x^2 + y^2)^2} \\ & = \frac{x^2 + y^2 - 2x^2}{(x^2 + y^2)^2} \\ & = \frac{y^2 - x^2}{(x^2 + y^2)^2}. \end{align*} Also, \begin{align*} \frac{\partial^2 u}{\partial x^2} & = \frac{\partial}{\partial x} \left( \frac{y^2 - x^2}{(x^2 + y^2)^2} \right) \\ & = \frac{(x^2 + y^2)^2(-2x) - (y^2 - x^2)2(x^2 + y^2)(2x)}{(x^2 + y^2)^4} \\ & = \frac{2x^3 - 6xy^2}{(x^2 + y^2)^3}. \end{align*}

Similarly, we have \begin{align*} \frac{\partial u}{\partial y} & = \frac{\partial}{\partial y} \left( \frac{x}{x^2 + y^2} \right) \\ & = \frac{-x(2y)}{(x^2 + y^2)^2} = \frac{-2xy}{(x^2 + y^2)^2} , \end{align*} and \begin{align*} \frac{\partial^2 u}{\partial y^2} & = \frac{\partial^2 }{\partial y^2} \left( \frac{-2xy}{(x^2 + y^2)^2} \right) \\ & = \frac{(x^2+y^2)^2(-2x) - (-2xy)2(x^2 + y^2)(2y)}{(x^2 + y^2)^4}\\ & = \frac{6xy^2 - 2x^3}{(x^2 + y^2)^3} . \end{align*} Therefore, \begin{align*} \frac{\partial ^2u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} = 0, \end{align*} and hence, $u$ satisfies the Laplace's equation.