I am attempting to solve this question from do Carmo's Riemannian Geometry. He begins by definining a Killing vector field in the following way:
Let $M$ a Riemannian manifold, $X$ a vector field on $M$. Let $p\in U \subset M$ be given. Let $\varphi : (\varepsilon,\varepsilon) \times U \to M$ be a differentiable mapping such that for all $q \in U$: \begin{align} &\varphi(0,q) = q \\ &\frac{\partial}{\partial t}\varphi(t,q)\bigg|_{t=0}=X(q)\\ &\varphi_t(q) \text{ is an isometry for all } t \in(-\varepsilon,\varepsilon) \end{align} Where $\varphi_t(q) = \varphi(t,q): U \to M$ for some fixed $t$.
On the problem I am stuck on, I am supposed to show that $X$ is a Killing vector field if and only if it satisfies the Killing equation: $$\langle\nabla_Y X,Z\rangle + \langle\nabla_Z X, Y\rangle = 0$$
The hint given in the forwards direction is to consider a Killing field $X$ and for any given $q\in M$ with $X(q) \neq 0$ a submanifold $S$ passing through $q$ normal to $X(q)$. Then $S$ has dimension $\dim M-1$ parametrized by coordinates $(x_1,\dots,x_{n-1})$. We can then parametrize a neighborhood of $q$ by $(x_1,\dots,x_{n-1},t)$ on $V \times (-\varepsilon,\varepsilon) \to M$ with $V \subset S$. Then for any $\partial_i=\frac{\partial}{\partial x_i}$ and $X(q) = \frac{\partial}{\partial t}$:
\begin{align} \langle\nabla_{\partial_i} X(q),\partial_j\rangle + \langle\nabla_{\partial_j} X(q), \partial_i\rangle &= X(q)\langle\partial_i,\partial_j\rangle - \langle[X(q),\partial_i] ,\partial_j\rangle-\langle[X(q),\partial_j] ,\partial_i\rangle\\ &= X(q)\langle\partial_i,\partial_j\rangle\\ &= \frac{\partial}{\partial t}\langle\partial_i,\partial_j\rangle \end{align}
Where the first equality is by compatibility with the metric and second equality is by symmetry of the connection. According to the hint, there should be a way to conclude that $\frac{\partial}{\partial t}\langle\partial_i,\partial_j\rangle=0$ using the fact that $\varphi_t(q)$ is an isometry, but I am not sure how to show that.