 Solution

Adopt the printed Riemann curvature tensor convention and contract $R_{ab}=R^c{}_{acb}$. At an arbitrary point choose normal coordinates for the original metric tensor. There the original connection coefficients vanish. Varying the coordinate curvature formula leavesAt that point these partial derivatives equal the covariant derivatives of the tensor $C$. Both sides are tensors, so the result is valid in every coordinate system:Contracting its first and third indices proves the Palatini identityThe normal-coordinate argument is applied independently at every point; it does not assume a flat background or set derivatives of the original connection to zero.