Harmonic function composed with conformal map is harmonic (in $\mathbb{R}^n$)
Though this is an old question, it seems worthwhile to set this matter straight:
The proof by @alext87 is flawed: it does not account the dependency of $x_\epsilon$ on $x$.
The statement is false in dimensions $n>2$: the composition of a harmonic function with a Möbius transformation is not harmonic in general. Example in 3 dimensions: let $u(x)=x_1$, which is harmonic. Let $v(x)=x/|x|^2$ be the inversion in the unit sphere, which is conformal. Now $(u\circ v)(x)=x_1/|x|^2$, and computation yields $\Delta(u\circ v)=-2x_1/|x|^4$.
The proof for $n=2$ can be found in many places, e.g., in Composition of a harmonic function with a holomorphic function
One can retain some of the connection between conformality and harmonicity in higher dimension by replacing the Laplacian $\Delta u=\mathrm{div}\,\nabla u$ with the $n$-Laplacian $\Delta_n u=\mathrm{div}\,(|\nabla u|^{n-2}\nabla u)$. The composition of an $n$-harmonic function with a Möbius transformation is $n$-harmonic.