I have been using the big O notation for some time now and know that, for example, $(e^x)^2 \in o(e^{x^2})$. In general, if $f$ grows faster than $g$, then $f(g(x))$ grows faster than $g(f(x))$. Does this hold in general? If so, how can I prove this? I was not able to find such proof anywhere. If not, does something similar hold, which would imply examples such as the one mentioned?
The version with "$f(g(x))$ grows strictly faster than $g(f(x))$" clearly does not hold (as correctly pointed out by Brian). Does the non-strict version hold?
It definitely does not hold in general. Let $f(x)=x^2$ and $g(x)=x$; then
$$\lim_{x\to\infty}\frac{g(x)}{f(x)}=\lim_{x\to\infty}\frac{x}{x^2}=0\;,$$
so $g(x)\in o(f(x))$, but $g\circ f=f\circ g$.
Taking $f(x)=1$ and $g(x)=\frac1x$ yields a similar example.