Is there some formula to determine which composition of functions grows faster than another without going to the limit of the ratio for $x\to\infty$ (but of course knowing what relationship there is between non-composition of them)?
I did some comparisons with the main functions (mainly exponential, logarithm, powers and factorials) and put them in order from the one with the highest order of infinity to the lowest
$x^{x}!\qquad\sqrt[x]{x^{x}!}\qquad x^{x^x}\qquad e^{x^x}\qquad x!^{x!}\qquad(x!)!\qquad \sqrt[x]{x!^{x!}}\qquad x^{x!}\qquad\sqrt[x]{(x!)!}\qquad ...$
$\ln(x)^{x!}\qquad\sqrt[x]{x^{x!}}\qquad \sqrt[x]{e^{x}!}\qquad \sqrt[x]{\ln(x)^{x!}}\qquad e^{x!}\qquad\sqrt[x]{e^{x!}}\qquad e^{x}!\qquad\sqrt[x]{e^{x}!}\qquad x!^{x}\qquad...$
$x!^{\ln(x)}\qquad \ln(x!)!\qquad x^{x}\qquad x!\qquad \ln(x)^x\qquad e^x\qquad x^{\ln(x)}\qquad \sqrt[x]{\ln(x!)!}\qquad\sqrt[x]{x!^{\ln(x)}}\quad...$
$\ln(x)^{\ln(x)}\qquad\ln(x)!\qquad\ln(x!)\qquad x^a\qquad\sqrt[x]{x!}\qquad\ln(\ln(x))!\qquad\ln(x)\qquad\ln(\ln(x))\qquad a$
Is there some way to determinate these order of approximation with some formula
This could be useful for determining simpler approximations for some functions and in computational complexity.
An example is the Stirling's approximation $x!\sim\sqrt{2\pi x}\left(\dfrac{x}{e}\right)^x$
I was thinking of some kind of formula that maps a real number to each function.
It suffices to use just the stirling's approximation, the exponential function and the logarithmic function. You make the approximation to a final function of form $f_1f_2(...(f_n(x))..)$. The most outer functions $f_1, f_2,...$ should be only $e^x$ or $\ln(x)$, the most inner function may have the form $x^\alpha (\ln(x))^{\beta}$.
Take for example the function $\sqrt[x]{\ln(x)^{x!}}$ which seems sufficiently complexe enough, we have: $$\begin{align} \sqrt[x]{\ln(x)^{x!}}&=\exp\left( (x-1)!\cdot\ln(\ln(x)) \right)\\ &\approx\exp\left( \sqrt{2\pi (x-1)}\cdot\left(\frac{x-1}{e}\right)^{x-1}\cdot\ln(\ln(x)) \right)\\ &\approx\exp\left( e^{\ln(\sqrt{2\pi})} e^{\frac{1}{2}\ln(x)} \cdot e^{(x-1)\ln(x-1)-(x-1)}\cdot e^{\ln(\ln(\ln(x)))} \right)\\ &\approx\exp\left( e^{\ln(\sqrt{2\pi})+\frac{1}{2}\ln(x) +(x-1)\ln(x-1)-(x-1)+\ln(\ln(\ln(x)))} \right)\\ &\approx\exp\left( e^{x\ln(x)} \right)=e^{e^{x\ln(x)}}\\ \end{align}$$
The other functions can be approximated easily by the same technique.