$M$ is a Riemannian manifold, and $V$ is a killing field, meaning its flow is a one parameter group of isometries. Then we have $\langle \dot \gamma(t), V(\gamma(t))\rangle$ is independent $t$ if $\gamma$ is a geodesic.
As usual, I take derivative of the desired expression: $\frac{d}{dt} \langle \dot \gamma(t), V(\gamma(t))\rangle = \langle \dot \gamma, \frac{D}{dt}V(\gamma(t))\rangle$. In order to show it is independent of $t$, I need the RHS to be zero, but I do not know how to proceed. Maybe calculating in corrdinates?
Use the equation $$\langle \nabla_Y X, Z \rangle + \langle Y,\nabla_Z X\rangle = 0,$$ where $X$ is a Killing vector field and $Y$ and $Z$ are arbitrary vectors. See wiki.
In particular, when $Y=Z$, by symmetry of the metric we end up with $$\langle Y, \nabla_Y X \rangle = 0.$$
Then use the fact that $\frac{D}{dt}V(\gamma(t)) = \nabla_{\dot{\gamma}}V$ when the vector field along a curve is induced by a smooth vector field on the manifold to conclude $$\langle \dot{\gamma}, \frac{D}{dt} V(\gamma(t)) \rangle = 0,$$ as desired.