Use
$$
[\nabla_\mu,\nabla_\nu]V^\rho=R_{\mu\nu}{}^{\rho\sigma} V_\sigma
$$
to conclude that
$$
\begin{aligned}
{}[\nabla^\nu\nabla_\nu,\nabla_\mu]\phi&\overset{ \mathrm A}=\nabla^\nu[\nabla_\nu,\nabla_\mu]\phi+[\nabla^\nu,\nabla_\mu]\nabla_\nu\phi\\
&\overset{ \mathrm B}=0+R_{\mu\nu}{}^{\nu\sigma} \nabla_\sigma\phi\\
&\overset{ \mathrm C}=R_{\mu\nu}\nabla^\nu\phi
\end{aligned}
$$

Note that in $\mathrm A$ we have used the fact that covariant derivatives commute with contractions, in $\mathrm B$ we have used $[\nabla_\nu,\nabla_\mu]\phi=[\partial_\nu,\partial_\mu]\phi=0$ (assuming a torsion-free connection), and in $\mathrm C$ we have used the definition of $R_{\mu\nu}$ as the contraction of the Riemann tensor.

This post imported from StackExchange MathOverflow at 2017-07-11 20:37 (UTC), posted by SE-user AccidentalFourierTransform