$\frac 1 {\left\| |y| x - \frac y {|y|} \right\|} = f(x)$ harmonic on the unit ball

Question

I got the following function which is supposed to be harmonic on the unit ball

Is there a quick way to show that this is true ?

$\frac 1 {\left\| |y| x - \frac y {|y|} \right\|} =f(x)$ for all $x,y \in $ the unit ball $\mathbb B_1(0) \subset \mathbb R^3$

I tried to find some "clever "substitution to solve this without too much calculation, but couldn't find any

Edit: $\Delta_x f(x,y)=0$ (only harmonic with respect to $x$)

Answer 1

Introduce $ z = \frac{y}{\Vert y \Vert^2} \in \mathbb{R}^3 \setminus \overline{\mathbb{B}_1 (0)}$ and write $$ f(x) = \frac{1}{\Vert y \Vert \Vert x -z \Vert} \equiv \frac{g(x-z)}{\Vert y \Vert} \, . $$ $g : \mathbb{R}^3 \setminus \{0\} \to \mathbb{R}, g(v) = \Vert v \Vert^{-1} ,$ is a smooth, harmonic function, which can be shown by direct calculation.

Thus $f : \mathbb{R}^3 \setminus \{z\} \to \mathbb{R}$ is smooth and harmonic by the chain rule. Since $z \not\in \mathbb{B}_1 (0)$, $f$ is harmonic on the unit ball.