I have, probably the most dumb question ever:
if we have $f = f(x)$ can we take derivative: $$\frac{dx}{df} = \frac{dx}{df(x)} ?$$
I was thinking of that it may be possible, if we find some inverse, or smth like substitution. So if $f= cx^2$, then: $$\frac{dx}{df} = \frac{dx}{dcx^2} $$ is actually: $\sqrt{c}x = \sqrt{y}$, and $x = \sqrt{\frac{y}{c}}$ then: $$\frac{dx}{df} = \frac{dx}{dcx^2} = \frac{d \sqrt{\frac{y}{c}} }{d y} =\sqrt{\frac{1}{c}} \frac{d \sqrt{y} }{d y} = ...$$ Is it a correct approach?
Suppose that $f$ is a function $D\to\Bbb R$. To make sense of what you write, we need that $f$ has an inverse function, that is, that there exists $g$ such that for all $x\in D$, $g(f(x)) = x$. Then using the chain rule, we have that $\frac{d(g\circ f)}{dx}(x) = \frac{dg}{df}(f(x))\times \frac{df}{dx}(x)$ for all $x\in D$. Note that since $g\circ f$ is the identity function, we have that the left-hand side of the equality is equal to 1 at every $x\in D$. Therefore, we have $\frac{dg}{df}(f(x)) = \frac1{\frac{df}{dx}(x)}$. Replacing $f(x)$ by $y$, this gives the formula $$\frac{dg}{df}(y) = \frac1{\frac{df}{dx}(g(y))}$$ and by abuse of notation, this quantity could also be written as $\frac{dx}{df}$.
Taking your example, if $f(x)=cx^2$ over $[0,+\infty]$, then $g(x) = \sqrt{\frac{x}{c}}$. We then have $\frac{dg}{df}(y) = \frac1{2\sqrt{cx}}$.