Answer
Using the Product Rule we show that
${f_{rr}} + \frac{1}{r}{f_r} = {r^{ - 1}}\frac{\partial }{{\partial r}}\left( {r\frac{{\partial f}}{{\partial r}}} \right)$
If $f$ is a radial harmonic function, using Eq. (13) we obtain the relation:
$r\frac{{\partial f}}{{\partial r}} = C$, where $C$ is constant.
Integrating we obtain
$f\left( {x,y} \right) = C\ln r + b$ for some constant $b$.
Work Step by Step
Using the Product Rule we evaluate
$\frac{\partial }{{\partial r}}\left( {r\frac{{\partial f}}{{\partial r}}} \right) = \frac{{\partial r}}{{\partial r}}\frac{{\partial f}}{{\partial r}} + r\frac{{{\partial ^2}f}}{{\partial {r^2}}} = {f_r} + r{f_{rr}}$
Multiply both sides by $\frac{1}{r}$ gives
$\frac{1}{r}\frac{\partial }{{\partial r}}\left( {r\frac{{\partial f}}{{\partial r}}} \right) = \frac{1}{r}{f_r} + {f_{rr}}$
Hence,
${f_{rr}} + \frac{1}{r}{f_r} = {r^{ - 1}}\frac{\partial }{{\partial r}}\left( {r\frac{{\partial f}}{{\partial r}}} \right)$
Suppose that $f$ is a radial function, then $f = f\left( r \right)$. It follows that ${f_\theta } = 0$. So, ${f_{\theta \theta }} = 0$.
Thus, Eq. (13) becomes
$\Delta f = {f_{rr}} + \frac{1}{{{r^2}}}{f_{\theta \theta }} + \frac{1}{r}{f_r} = {r^{ - 1}}\frac{\partial }{{\partial r}}\left( {r\frac{{\partial f}}{{\partial r}}} \right)$
If $f$ is also harmonic, then $\Delta f = {r^{ - 1}}\frac{\partial }{{\partial r}}\left( {r\frac{{\partial f}}{{\partial r}}} \right) = 0$. This implies that $r\frac{{\partial f}}{{\partial r}}$ is constant.
Write $r\frac{{\partial f}}{{\partial r}} = C$, where $C$ is constant.
Hence, if $f$ is a radial harmonic function, then $r{f_r} = C$ for some constant $C$.
Now, we can write $r{f_r} = C$ as $r\frac{{df}}{{dr}} = C$, since $f$ is a function of $r$ alone. So,
$df = \frac{C}{r}dr$
Integrating gives $f\left( r \right) = C\ln r + b$, where $b$ is an integration constant.
Hence, we conclude that $f\left( {x,y} \right) = C\ln r + b$ for some constant $b$.
