Let $f(x,y,z)=\ln{(x^2+y^2+z^2)}$. Then we have: \begin{align} f_x(x,y,z)&=\frac{d}{dx}\left(f(x,y,z)\right)=\frac{d}{dx}(\ln{(x^2+y^2+z^2)})\\ &=\frac{1}{x^2+y^2+z^2}\left(\frac{d}{dx}(x^2+y^2+z^2)\right)=\frac{2x}{x^2+y^2+z^2}. \end{align} Further, we have that: \begin{align} f_{xx}(x,y,z)&=\frac{d}{dx}\left(\frac{2x}{x^2+y^2+z^2}\right)\\ &=\frac{(2x)'(x^2+y^2+z^2)-(x^2+y^2+z^2)'(2x)}{(x^2+y^2+z^2)}\\ &=\frac{2x^2+2y^2+2z^2-4x^2}{(x^2+y^2+z^2)^2}=\frac{-2x^2+2y^2+2z^2}{(x^2+y^2+z^2)^2}. \end{align} By symmetry, we have that $f_{yy}(x,y,z)=\dfrac{2x^2-2y^2+2z^2}{(x^2+y^2+z^2)^2}$ and $f_{zz}(x,y,z)=\dfrac{2x^2+2y^2-2z^2}{(x^2+y^2+z^2)^2}$. Thus, \begin{align*} f_{xx}(x,y,z)+f_{yy}(x,y,z)+f_{zz}(x,y,z)&=\dfrac{-2x^2+2y^2+2z^2+2x^2-2y^2+2z^2+2x^2+2y^2-2z^2}{(x^2+y^2+z^2)^2}\\ &=\frac{2x^2+2y^2+2z^2}{(x^2+y^2+z^2)^2}=2\frac{x^2+y^2+z^2}{(x^2+y^2+z^2)^2}=\frac{2}{x^2+y^2+z^2}. \end{align} Since $f_{xx}+f_{yy}+f_{zz}\neq 0$, $f$ is not harmonic.