I'm trying to prove the Ricci identity
Let $Z^a$ be a vector field, $R^a_{\,bcd}$ the Riemann curvature tensor and $\nabla$ a torsion-free connection. Then: $\nabla_c\nabla_dZ^a-\nabla_d\nabla_cZ^a=R^a_{\,bcd}Z^b$.
In particular, I want to start from the RHS. To do so, I've multiplied it by two arbitrary vector fields $X^c$ and $Y^d$, and used the definition of the Riemann tensor:
$$\begin{align}R^a_{\,bcd}Z^bX^cY^d&=\left(R(X,Y)Z\right)^a\\ &=\left(\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z\right)^a\\ &=\left(\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\partial_{[X,Y]}Z\right)^a \end{align}$$
where in the last line I use the fact that the connection is torsion-free to move from covariant to partial derivative. Obviously, the next step is showing that the partial derivative of $Z$ with respect to $[X,Y]$ vanishes.
If they were basis vectors, I know that $[e_\mu,e_\nu]=0$, but what about the case of general vector fields?
EDIT: second attempt
$$\begin{align}R^a_{\,bcd}Z^bX^cY^d&=\left(R(X,Y)Z\right)^a\\ &=\left(\nabla_X\nabla_YZ-\nabla_Y\nabla_XZ-\nabla_{[X,Y]}Z\right)^a\\ &=\left(X^c\nabla_c\left(Y^d\nabla_dZ\right)-Y^d\nabla_d\left(X^c\nabla_cZ\right)-\nabla_{X^c\nabla_cY}Z+\nabla_{Y^d\nabla_dX}Z\right)^a\\ &=\left(\left(X^c\nabla_cY\right)^d\nabla_dZ+X^cY^d\nabla_c\nabla_dZ-\left(Y^d\nabla_dX\right)^c\nabla_cZ-Y^dX^c\nabla_d\nabla_cZ\right.\\ &\left.\qquad-\left(X^c\nabla_cY\right)^d\nabla_dZ+\left(Y^d\nabla_dX\right)^c\nabla_cZ\right)^a\\ &=\left(X^cY^d\nabla_c\nabla_dZ-Y^dX^c\nabla_d\nabla_cZ\right)^a\\ &=X^cY^d\left(\nabla_c\nabla_d-\nabla_d\nabla_c\right)Z^a \end{align}$$