I am dealing with a problem where I have the derivative $f'(x)$ in function of its antiderivative $f(x)$. How can I solve this?
$f'(f(x))$ is linear (just assume any linear function, for sake of simplicity), but I almost surely know that $f(x)$ is a hyperbole.
@edit: The problem is along the lines of: "A particle travels towards a point $r$ such that, in any instant, the remaining time until its arrival $q$ with its instant velocity $V(S)$ is the same. Find $S(t)$."
From there it was easy to deduce the following:
$$q = \frac{r - S}{V(S)}$$
$$V(S) = \frac{r - S}{q}$$
$$V(S(t)) = \frac{r - S(t)}{q}$$
I tried to solve it in a physics way, using classical dynamics, but the results were weird. Is there any other way to solve it?
I think you're getting yourself led astray by bad notation -- especially, you're writing $V(S(t))$ where you should just be writing $V(t)$ or $\frac{d}{dt}S(t)$ or $S'(t)$. The problem text you're quoting seems to use the same wrong notation, so it's not purely your fault, but what you should be doing is $$ q = \frac{r-S(t)}{S'(t)} $$ rearranging to $$ S'(t) = \frac 1q(r-S(t)) $$
If we switch to a new variable $u(t)=r-S(t)$ we get $$ u'(t) = \frac{-1}q u(t) $$ which you ought to recognize as the standard differential equation whose solution is $$ u(t) = C e^{-t/q} $$ where $C$ is a constant of integration, and therefore $$ S(t) = r - C e^{-t/q} $$
In other words, the distance to point $r$ varies as an exponential decay with time constant $q$. (With a bit of experience you will be able to recognize that directly from the problem description, and not need to do the algebra explicitly).