Reparametrised curve is geodesic I do not know how to continue with my exercise:
I have a pseudo-Riemannian manifold $(M,g)$, $I,J \subset \mathbb{R}$ open intervals and $\gamma:I \rightarrow M$ a smooth curve, $\gamma' \neq 0$. Also there exists $f \in C^{\infty}(M)$, s.t. $\nabla_{\gamma'}\gamma'=f\gamma'$
Now let $h:J \rightarrow I$ be a diffeomorphism, s.t. for the curve $\tilde{\gamma}:J \rightarrow M, \tilde{\gamma}=\gamma \circ h$ it holds $\langle\tilde{\gamma}', \tilde{\gamma}'\rangle \equiv c >0$. 
I have to show that $\tilde{\gamma}$ is a geodesic and that $h''(t)=h'(t)^2 f(h(t))$.
Now I know that:
$\nabla_{\tilde{\gamma}'} \tilde{\gamma}'= h'' \gamma' \circ h + (h')^2 \nabla_{\gamma' \circ h} \gamma' \circ h=(h''  +(h')^2(f \circ h))(\gamma' \circ h)$
So since $\gamma' \neq 0$ I can conclude the second part of the exercise if I can show that $\tilde{\gamma}$ is geodesic. But here, I don't know what to do, I've tried using the Koszul formula, where I can use that $\tilde{\gamma}$ has constant speed but it didn't work out..
 A: Assume that $\gamma :[0,1]\rightarrow M$ is a curve Then
$$
\gamma'(t)= \sum_i\ a_i(t)E_i(t) $$ where $E_i$ is a coordinate
field. Hence $$ \nabla_{\gamma'(t)} \gamma'(t)= \frac{d}{dt} a_i(t)
E_i(t) + a_i(t) a_j(t) \nabla_{E_j} E_i $$
When $f\gamma '=\nabla_{\gamma'(t)} \gamma'(t)$, then $$ fa_i =
\frac{d}{dt} a_i + a_la_m\Gamma_{lm}^i $$
When $a =\gamma\circ h$ has unit speed with $h(s)=t$, then $$ a'(s)=
h'(s) \gamma'(h(s))= h'(s) a_i(h(s)) E_i(a(s)) $$
Hence \begin{align*} \nabla_{a'(s)} a'(s)& =\frac{d}{ds} \{ h' a_i\circ h \} E_i
+h' a_l h' a_m \Gamma_{lm}^i E_i \\& =( h')^2 \{ \frac{d}{dt} a_i + a_la_m \Gamma_{lm}^i \} E_i
\\& =(h')^2 fa_iE_i =C a' \end{align*}
 for some function $C$. Since $a'(s)$ is unit speed, then $C=0$ 

OLD : When $\nabla_{c'}c' =fc'$ and $a=c\circ h$ is unit speed, then $$
 a' = h' c',\ \nabla_{a'}a' =\nabla_{h'c'}(h'c')
 =h' c'(h') c' +(h')^2 fc'  = \underbrace{\{ c' (h') + f
 h'\}}_{=F}
 a' $$
Note that $0=a'(a',a')=2(\nabla_{a'} a',a')
 =2F $ so that $F=0$
Add : We follow the notation in OP :
$\nabla_{\overline{\gamma}'} \overline{\gamma}'= \nabla_{h'\gamma '}
h'\gamma' =h'\gamma'(h')\gamma' + (h')^2 f\gamma' $
Assume that $\nabla_{\overline{\gamma}'} \overline{\gamma}' =0,\
\gamma'\neq 0,\ h'\neq 0 $ Then
$$ \gamma'(t) (h'(s)) + h'f =0 $$
When $t=h(s),\ s=Q(t),\ h\circ Q (t)=t$, then
$$ h'Q'=1,\ h'f=-\frac{d}{dt} h'(s)=-h'' \frac{d}{dt} s=
-h'' \frac{1}{h'} $$
A: We have $\nabla_{\tilde\gamma'}\tilde\gamma'\perp\tilde\gamma'$ from differentiating $\langle\tilde\gamma',\tilde\gamma'\rangle=c$, but also $\nabla_{\tilde\gamma'}\tilde\gamma'\parallel\tilde\gamma'$ from $\nabla_{\gamma'}\gamma'\parallel\gamma'$.
