Locally symmetric spaces and the curvature tensor 
Let $M$ be a Riemannian manifold. Suppose $\nabla R=0$ where $R$ is the curvature tensor (we then say $M$ is locally symmetric). Then if $\gamma$ is a geodesic of $M$ and $X,Y,Z$ are parallel vector fields along $\gamma$ then $R(X,Y)Z$ is a parallel field along $\gamma$.

However, I do not fully understand what does it mean $\nabla R=0$. Is it to say that $\nabla_ZR(X,Y,W,T)=0$ for all vector fields $Z,X,Y,W,T$?
How can we use this to prove $\dfrac{D}{dt}R(X,Y)Z=0$?
The only thing I can think is maybe we can use that if $V(t)=Y(c(t))$ (i.e. $V$ is induced by a vector field) then $\dfrac{DV}{dt}=\nabla_{c'}Y$.
Any hints would be appreciated.

The covariant differential $\nabla T$ of a tensor $T$ of order $r$ is defined as:
$$\nabla T (Y_1, \ldots, Y_r, Z) = Z(T(Y_1, \ldots, Y_r)) - T(\nabla_Z Y_1, \ldots, Y_r) - \cdots - T(Y_1, \ldots, Y_{r-1}, \nabla_Z Y_r)$$
In this case $T = R$ and $r = 4$ because the Riemann tensor takes 4 arguments $R(X,Y,Z,W)$. So we have
$$(\nabla R)(X,Y,Z,W,V)=V(R(X,Y,Z,W))-R(\nabla_V X, Y, Z, W)-R(X,\nabla_VY,Z,W)-R(X,Y,\nabla_VZ,W)-R(X,Y,Z,\nabla_VW)$$
