Answer
Using Eq. (13) we show that $\Delta f = 0$. Hence, by definition $f$ is harmonic.
Work Step by Step
We have $f\left( {x,y} \right) = \ln r$.
The partial derivatives with respect to $r$ and $\theta$ are
$\frac{{\partial f}}{{\partial r}} = \frac{1}{r}$, ${\ \ \ }$ $\frac{{\partial f}}{{\partial \theta }} = 0$
$\frac{{{\partial ^2}f}}{{\partial {r^2}}} = - \frac{1}{{{r^2}}}$, ${\ \ \ }$ $\frac{{{\partial ^2}f}}{{\partial {\theta ^2}}} = 0$
Using Eq. (13) we obtain
$\Delta f = {f_{rr}} + \frac{1}{{{r^2}}}{f_{\theta \theta }} + \frac{1}{r}{f_r}$
$\Delta f = - \frac{1}{{{r^2}}} + \frac{1}{r}\cdot\frac{1}{r} = 0$
Since $\Delta f = 0$, by definition $f$ is harmonic.
