Question: In q. 22 in page 141, I am asked to show that if 

$$U^{\alpha}\nabla\_{\alpha} V^{\beta} = W^{\beta},$$ 

then 

$$U^{\alpha}\nabla\_{\alpha}V\_{\beta}=W\_{\beta}.$$

Here's what I have done:
$$V\_{\beta}=g\_{\beta \gamma} V^{\gamma},$$
so
$$U^{\alpha} \nabla\_{\alpha} (g\_{\beta \gamma} V^{\gamma})=U^{\alpha}(\nabla\_{\alpha} g\_{\beta \gamma}) V^{\gamma} + g\_{\beta \gamma} (U^{\alpha} \nabla\_{\alpha} V^{\gamma}).$$

Now, I understand that the second term is $W\_{\beta}$, but how come the first term vanishes?

Answer: 
The covariant derivative is metric compatible, so $\nabla\_{\alpha} g\_{\beta \gamma} = 0$. This is the condition that the inner product is preserved under parallel transport.
I don't think you need metric compatibility to prove this although you can use it. There is a much simpler way with repeated use of (any) metric to lower the index.

$
U^\alpha\nabla\_\alpha V^\beta=W^\beta
$

$
\Rightarrow U^\alpha g^{\beta\gamma}\nabla\_\alpha V\_\gamma=g^{\beta\gamma}V\_\gamma
$

$
\Rightarrow U^\alpha g\_{\mu\beta}g^{\beta\gamma}\nabla\_\alpha V\_\gamma=g\_{\mu\beta}g^{\beta\gamma}V\_\gamma
$

$
\Rightarrow U^\alpha\delta\_\mu^\gamma\nabla\_\alpha V\_\gamma=\delta\_\mu^\gamma V\_\gamma
$

$
\Rightarrow U^\alpha\nabla\_\alpha V\_\mu=V\_\mu
$

there are three ways to do the first step, only one uses metric compatibility. More at <https://www.general-relativity.net/2019/10/symmetries-and-killing-vectors.html>