No one answered this week's question. You can find my solution below.
[sp]Proof: ($\Leftarrow$): Assume that $\dfrac{\,d}{\,dt}\left<V,W\right> =\left<\dfrac{\,DV}{\,dt},W\right>+\left<V,\dfrac{\,DW}{\,dt}\right>$ is true. Now suppose that $V$ and $W$ are parallel vector fields along the curve $c$. By definition of parallelism, $\dfrac{\,DV}{\,dt}=\dfrac{\,DW}{\,dt}=0$. Therefore,$$\frac{\,d}{\,dt}\left<V,W\right>=\left<0,W\right>+\left<V,0\right>=0.$$
This implies that $\left<V,W\right>$ is constant.
($\Rightarrow$): Suppose that $\nabla$ is compatible with $\left<\cdot,\cdot\right>$. For some point $t_0$ on the curve $c\left(t\right)$, let us define an orthonomal basis $\{E_1\!\left(c\left(t_0\right)\right),E_2\!\left(c\left(t_0\right)\right),\ldots,E_m\!\left(c\left(t_0\right)\right)\}$ of the tangent space $T_{c\!\left(t_0\right)}M$. By parallel transport (proposition 4.2.5), we can extend these vectors to the parallel vector fields $\{E_1\!\left(c\left(t\right)\right), E_2\!\left(c\left(t\right)\right), \ldots,E_m\!\left(c\left(t\right)\right)\}$. Since parallel transport preserves the lengths of these basis vectors, $\{E_1\!\left(c\left(t\right)\right), E_2\!\left(c\left(t\right)\right), \ldots,E_m\!\left(c\left(t\right)\right)\}$ is now an orthonormal basis for $T_{c\!\left(t\right)}M$ for any $t\in I$. Take $V=\displaystyle\sum_{i=1}^m v_iE_i$ and $W=\displaystyle\sum_{j=1}^m w_jE_j$. By definition of the Riemannian metric and noting that $\left<E_i,E_j\right>=1$, we see that
$$\begin{aligned}\left<V,W\right>&=\left<\sum_{i=1}^m v_iE_i,\sum_{i=1}^m w_jE_j\right>\\ &= \sum_{i=1}^m v_iw_i.\end{aligned}$$
Therefore, $\dfrac{\,d}{\,dt}\left<V,W\right>= \displaystyle\sum_{i=1}^m\left(\frac{\,dv_i}{\,dt}w_i +v_i\frac{\,dw_i}{\,dt}\right)$
We now verify the RHS of the equation is true.
$$\begin{aligned}\left<\frac{\,D_cV}{\,dt},W\right>&=\left<\sum_{i=1}^m\frac{D\!\left(v_iE_i\right)}{\,dt},W\right>\\ &= \left<\sum_{i=1}^m\frac{\,dv_i}{\,dt}E_i+v_i\frac{DE_i}{\,dt},\sum_{j=1}^mw_jE_j\right>\\ &= \left<\sum_{i=1}^m\frac{\,dv_i}{\,dt}E_i,\sum_{i=1}^m w_iE_i\right>\\&=\sum_{i=1}^m\frac{\,dv_i}{\,dt}w_i.\end{aligned}$$
Similarly,
$$\begin{aligned}\left<V,\frac{\,D_cW}{\,dt}\right>&=\left<V,\sum_{j=1}^m\frac{D\!\left(w_jE_j\right)}{\,dt}\right>\\ &= \left<\sum_{i=1}^mv_iE_i,\sum_{j=1}^m\frac{\,dw_j}{\,dt}E_j+w_j\frac{DE_j}{\,dt}\right>\\ &= \left<\sum_{i=1}^mv_iE_i,\sum_{i=1}^m\frac{\,dw_i}{\,dt}E_i\right>\\&=\sum_{i=1}^m\frac{\,dw_i}{\,dt}v_i.\end{aligned}$$
We now see that
$$\begin{aligned}\frac{\,d}{\,dt}\left<V,W\right>&=\sum_{i=1}^m\frac{\,dv_i}{\,dt}w_i+v_i\frac{\,dw_i}{\,dt}\\ &= \sum_{i=1}^m\frac{\,dv_i}{\,dt}w_i+\sum_{i=1}^mv_i\frac{\,dw_i}{\,dt}\\ &= \left<\frac{D_cV}{\,dt},W\right>+\left<V,\frac{\,D_cW}{\,dt}\right>.\end{aligned}$$
This completes the proof.$\hspace{.25in}\blacksquare$[/sp]