First of all, I'm a beginner in calculus and this problem came out to be tricky for me.
Denote $\dot{x}=\frac{dx}{dt}$ for any function $x(t)$ of time $t$. Consider the equality
$$\dot{y}=c \frac{\dot{x}}{x},$$
where $y=:y(t)$ and $x=:x(t)$. My first question is: it holds that
$$\frac{d\dot{y}}{d\dot{x}}=c/x$$ or should I take into account that $\dot{y}=f(\dot{x},x)$ and $\dot{x}=g(x)$?
Moreover, I would like to know how to treat the second derivative
$$\frac{d^2\dot{y}}{d\dot{x}^2}.$$
Finally, for some function $h$, is it correct that $\frac{dh(x)}{d\dot{x}}=\frac{dh}{dx}\frac{dx}{d\dot{x}}$, provided that $\frac{d\dot{x}}{dx}=\frac{d\dot{x}}{dt}\frac{dt}{dx}=\frac{\ddot{x}}{\dot{x}}$ exists?
Thanks in advance!
In your first equation I assume you want to obtain something like $$\frac{dy}{dx}=\frac{\dot y}{\dot x}=\frac{c}x?$$ This assumes that $x$ can be taken as independent parameter. The $t-x$ relation can be inverted on monotonous segments of the function $x(t)$.
Assuming that $\dot x$ is (locally) suitable as independent parameter, you would get $t$ as a function of $\dot x$ and thus $$\frac{d\dot y(t(\dot x))}{d\dot x}=\ddot y\frac{dt}{d\dot x}$$ and on the other side it is equal to $$=\frac{d}{\dot x}\frac{c\dot x}{x(t(\dot x))}=\frac{c}x-\frac{c\dot x^2\frac{dt}{d\dot x}}{x^2}.$$ This does not look very close to what you wanted to achieve. After what you expected, there come some other term that expresses that $x$ is not constant along solution paths.