$\displaystyle Given\ u=\frac{1}{2}\log(x^2+y^2) \\[2ex] \displaystyle Differentiating\ w.r.t.\ \ \ x \\[2ex] \displaystyle Ux=\frac{1}{2}\bullet{}\ \frac{1}{x^2}+y^2.2x=\frac{x}{x^2}+y^2 \\[2ex] $

$\displaystyle Again,\ Partially\ differentiating\ w.r.t.x \\[2ex] \displaystyle Uxx=(x^2+y^2)\bullet{}1-x\bullet{}\frac{2x}{(x^2+{y^2)}^2}\ =\ y^2-\ \frac{x^2}{(x^2+{y^2)}^2}…\left(1\right) \\[2ex] $

$\displaystyle Similarly,\ differentiating\ ‘u’\ partially\ w.r.t.\ ‘y’ \\[2ex] \displaystyle Uy=\frac{1}{2}.\ \frac{1}{x^2}+y^2.2y=\frac{y}{x^2+y^2} \\[2ex] $

$\displaystyle Again,\ differentiating\ partially\ w.r.t.\ ‘y’ \\[2ex] \displaystyle Uyy=(x^2+y^2)\bullet{}1-y\bullet{}\frac{2y}{(x^2+{y^2)}^2}\ =\ x^2-\ \frac{y^2}{(x^2+{y^2)}^2}=\frac{{-(y}^2-\ x^2)}{(x^2+{y^2)}^2} \\[2ex] =\ -Uxx...\ from\ (1)\\[2ex]\therefore{}Uxx+Uyy=0,\ which\ is\ a\ laplace’s\ equation \\[2ex] \displaystyle \therefore{}\ ‘u’\ satisfies\ laplace’s\ equations,\ hence\ it\ is\ a\ harmonic\ function. \\[2ex]$