The coefficients in the spherical <Laplacian> are independent of $\phi$, so $[\Delta,\partial_\phi]=0$. Also $\partial_\phi$ is tangent to both boundary spheres; differentiating the zero boundary trace gives another zero trace. Therefore $w=\partial_\phi u$ satisfies $\Delta w=\partial_\phi f$ and the same homogeneous <Dirichlet boundary conditions>.

In Cartesian coordinates $\partial_\phi=x_1\partial_2-x_2\partial_1$, a globally smooth rotation field even where angular coordinates are singular. Permuting coordinate axes gives $R_{12}$, $R_{23}$ and $R_{31}$. For each, the <rotational commutator for the Laplacian> follows directly from $\Delta(x_i\partial_ju)=x_i\partial_j\Delta u+2\partial_i\partial_ju$; antisymmetrizing cancels the extra terms. Thus
$$
\boxed{\Delta(R_{ij}u)=R_{ij}f,\qquad R_{ij}u|_{\partial\Omega}=0.}
$$
If $f$ is radial, all these right-hand sides vanish. Uniqueness from the energy estimate makes every $R_{ij}u$ zero. At each point of a sphere the rotation fields span its tangent plane, so all tangential <derivatives> of $u$ vanish. Each sphere is connected, giving $\boxed{u(x)=U(|x|)}$. This proves radiality without assuming it as a separation-of-variables ansatz.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 105 / 2 / g / Solution

{title2=$[\Delta,x_i\partial_j-x_j\partial_i]=0$}

The infinitesimal rotation $R_{ij}=x_i\partial_j-x_j\partial_i$ commutes with the <Laplacian>: the extra terms $2\partial_i\partial_j$ cancel. It is tangent to every sphere, so it preserves zero Dirichlet traces on spherical shells. If $\Delta u=f$ and $f$ is radial, uniqueness applied to $R_{ij}u$ forces all these rotation <derivatives> to vanish, hence $u$ is radial.

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 Incoming links: Rotational commutator for the Laplacian