Show that $\log(\Vert \cdot\Vert)$ is harmonic

Question

I want to show that $g=\log(\Vert \cdot\Vert)\colon\mathbb{R}^2\setminus\{0\}\to\mathbb{R}$ is harmonic. It should be $\partial_j g(x)=\frac{x_j}{\Vert x\Vert^2}$, so I thought $\partial_j g= pr_j\cdot \frac{1}{\Vert\cdot\Vert^2}$, so by $$\partial pr_j(x)=pr_j,\quad \partial\frac{1}{\Vert \cdot\Vert}(x)=\frac{-2}{\Vert x\Vert^3}\frac{1}{\Vert x\Vert} \langle\,\cdot\,,x\rangle$$

and the product rule it should be $$\partial^2_jg(x)= \partial(\partial_jg)(x)e_j=\frac{1}{\Vert x\Vert^2}-\frac{2}{\Vert x\Vert^4}x_j.$$ This seems to be wrong, as, under the Laplace operator, g does not evaluate to zero. Where am I wrong?

Answer 1

$$\partial_j (\partial_jg(x))=\partial_j\frac{x_j}{\Vert x\Vert^2}=\frac{\Vert x\Vert^2-2x_j\cdot x_j}{\Vert x\Vert^4}=\frac{1}{\Vert x\Vert^2}-\frac{2x_j^2}{\Vert x\Vert^4}$$ by the quotient rule and taking the sum gives us $$\sum_{j=1}^2\partial_j (\partial_jg(x))=\frac{2}{\Vert x\Vert^2}-\frac{2\Vert x\Vert^2}{\Vert x\Vert^4}=0$$