I have the following ODE system and want to show that all solutions $\gamma(t)=(x(t),y(t),z(z))$ exist for all times (or can be extended on all of $\mathbb{R}$). The only problem is that the system depends on an unknown function $c:\mathbb{R}\rightarrow (0,\infty)$, which is smooth, positive and periodic.
\begin{align*} \ddot{x}(t)&=\frac{1}{2}\frac{c'(z(t))}{c(z(t))}\dot{y}(t)^2 \\ \ddot{y}(t)&=-\frac{c'(z(t))}{c(z(t))}\dot{y}(t)\dot{z}(t)\\ \ddot{z}(t)&=-\frac{c'(z(t))}{c(z(t))}\dot{z}(t)^2. \end{align*}
I can rewrite the last equation to get $c(z)\ddot{z}+c'(z)\dot{z}^2=0$, so $\frac{d}{dt}(\dot{z}c(z))=0$, which tells me that $\dot{z(t)}c(z(t))=a_1$ for some constant $a_1\neq 0$. Solving this gives the implicit expression \begin{align*} a_1 t+a_2=\int_{a_3}^{z(t)}c(\xi)d\xi \end{align*} for $a_2,a_3$ constants. Substituting $\dot{z}=\frac{a_1}{c(z)}$ in the second equation I get $\ddot{y}=-a_1\frac{c'(z)}{c(z)^2}\dot{y}$ which can also be written as $\ddot{y}=\frac{\ddot{z}}{\dot{z}}\dot{y}$. Now I don't know how to continue. I'm pretty sure one cannot proceed solving this without knowing what the function $c$ is. But since $c$ is periodic (so in particular $c$ and $c'$ are bounded) I have hope that one nevertheless can show that all solutions exist for all time.
Not a full answer, but this is too long to fit into a comment.
Write the second equation as
$$ \frac{\ddot y}{\dot y} = \frac{\ddot z}{\dot z} $$
and integrate both sides to get
$$ \ln \dot y = \ln \dot z + \ln b_0 \implies \dot y = b_0\dot z \implies y(t) = b_0z(t) + b_1 $$
where $b_0$, $b_1$ are arbitrary constants to be determined by initial conditions
The first equation is
$$ \ddot x = -\frac12 \frac{\ddot z}{\dot z^2}\dot y^2 = -\frac{b_0^2}{2}\ddot z \implies x(t) = -\frac{b_0^2}{2}z(t) + b_2t + b_3 $$
If you can show that there is a one-to-one correspondence between $z$ and $t$, then the solution exists.