= Solution
Use the <Levi-Civita connection> interpretation of the curvature convention. Its <Riemann curvature tensor> satisfies
$$
\boxed{R_{abcd}=-R_{bacd}=-R_{abdc},\qquad R_{abcd}=R_{cdab},\qquad R_{[abc]d}=0.}
$$
The first antisymmetry comes from the derivative commutator; the second comes from compatibility with the <metric tensor>. The cyclic identity is the <first Bianchi identity>, and together these imply exchange symmetry of the two antisymmetric index pairs. Contracting gives the symmetric <Ricci tensor>, $\boxed{R_{ab}=R_{ba}}$. The differential <Bianchi identity> is $\nabla_{[a}R_{bc]de}=0$.
There is a hypothesis qualification. A <metric connection> need not be a <torsion-free connection>. For a general such connection, the tensorial derivative commutator contains $-T^c{}_{ab}\nabla_c$ as well as curvature; pair exchange, the stated cyclic identity and Ricci symmetry need not hold. The printed pure-curvature commutator, and the scalar identity requested later, use the torsion-free interpretation. Under that intended interpretation all the boxed symmetries follow. They are not assertions about arbitrary metric-compatible connections with <torsion tensor>.
For example, with a constant Minkowski metric a connection having only $\Gamma^1{}_{02}=1$ and $\Gamma^2{}_{01}=-1$ nonzero is metric compatible, since its connection matrix in direction zero is a spatial rotation. It has $T^1{}_{02}=1$. For the scalar $f=x^1$ the tensorial Hessian commutator is $[\nabla_0,\nabla_2]f=-1$, rather than zero. This explicitly demonstrates the missing zero-torsion condition used in the subsequent scalar proof.
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