If $X$ is Killing and $p$ is a critical point of $f(p) = \|X(p)\|^2$ then the trajectory of $X$ through $p$ is geodesic.

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Let $(M,g)$ a Riemannian manifold and $X$ a Killing field on $M$. Then, we can define the function:

$$f(p) = \|X(p)\|^2$$

The claim is:

If $p$ is a critical point of $f$ then the flow of $X$ at $p$ is a geodesic.

My attempts:

$$Xf(p) = X(\|X(p)\|^2) = 2g(\nabla_XX(p),X(p)).$$

But then, once $p$ is a critical point then $$g(\nabla_XX(p),X(p)) = 0.$$

But if I am not mistaken the condition of $X$ Killing implies the same. Right?

How to proceed?

I must conclude that $\nabla_XX(p) = 0.$

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$0=\frac{1}{2}Vf=g(\nabla_VX,X)= -g(\nabla_XX,V)$ for any $V$

That is $\nabla_XX(p)=0$

As you pointed, $X|X|^2=0$ This implies that $|X|^2$ is constant along a flow That is if $p$ is critical, flow passing through $p$ is set of critical point