In this answer, we assume that $f:\mathbb{Q}_{>0}\to\mathbb{Q}_{>0}$. Let $P(x,y)$ denote the condition that
$$f\Big(y\,\big(f(x)\big)^2\Big)=x^3\,f(xy)\,.$$ First, we shall prove that $f$ is injective. If $x$ and $y$ are positive rational numbers such that $f(x)=f(y)$, then
$$x^3\,f(x)\overset{P(x,1)}{=\!=\!=}f\Big(\big(f(x)\big)^2\Big)=f\Big(\big(f(y)\big)^2\Big)\overset{P(y,1)}{=\!=\!=}y^3\,f(y)\,.$$
As $f(x)=f(y)>0$, we get $x^3=y^3$, whence $x=y$.
Observe that
$$\begin{align}f\Big(\big(f(x)\big)^2\,\big(f(y)\big)^2\Big)&\overset{P\Big(y,\big(f(x)\big)^2\Big)}{=\!=\!=\!=\!=\!=\!=}y^3\,f\Big(y\,\big(f(x)\big)^2\Big)
\overset{P(x,y)}{=\!=\!=}y^3\,\Big(x^3\,f(xy)\Big)
\\&=(xy)^3\,f(xy)
\overset{P(xy,1)}{=\!=\!=\!=}f\Big(\big(f(xy)\big)^2\Big)\,.\end{align}$$
Because $f$ is injective, we get $$\big(f(x)\big)^2\,\big(f(y)\big)^2=\big(f(xy)\big)^2\,.$$
This shows that $f$ is multiplicative, namely,
$$f(xy)=f(x)\,f(y)\text{ for all }x,y\in\mathbb{Q}_{>0}\,.$$
Thus, from $P(x,1)$, we get
$$\Big(f\big(f(x)\big)\Big)^2=f\left(\big(f(x)\big)^2\right)=x^3\,f(x)\text{ for every }x\in\mathbb{Q}_{>0}\,.$$
Define $g(x):=x\,f(x)$ for each $x\in\mathbb{Q}_{>0}$. Note that
$$\begin{align}\Big(f\big(g(x)\big)\Big)^2&=f\left(x^2\,\big(f(x)\big)^2\right)=\big(f(x)\big)^2\,\Big(f\big(f(x)\big)\Big)^2
\\&=\big(f(x)\big)^2\,\big(x^3\,f(x)\big)=\big(x\,f(x)\big)^3=\big(g(x)\big)^3\,.\end{align}$$
Consequently,
$$\Big(g\big(g(x)\big)\Big)^2=\Big(g(x)\,f\big(g(x)\big)\Big)^2=\big(g(x)\big)^2\,\big(g(x)\big)^3=\big(g(x)\big)^5\,.\tag{*}$$
We claim that $g\equiv 1$. To prove this, we define the height $\text{ht}(r)$ of a rational number $r>0$ to be the largest integer $\beta\geq 0$ such that $2^\beta$ divides $\alpha_1,\alpha_2,\ldots,\alpha_k$, if $$r=p_1^{\alpha_1}\,p_2^{\alpha_2}\,\cdots\,p_k^{\alpha_k}\,,$$
where $p_1,p_2,\ldots,p_k$ are pairwise distinct prime natural numbers and $\alpha_1,\alpha_2,\ldots,\alpha_k\in\mathbb{Z}$. We set $\text{ht}(1)$ to be $\infty$. Clearly, $\text{ht}\big(f(x)\big)\geq \text{ht}(x)$ due to multiplicativity of $f$.
To finish the proof, we assume on the contrary that $g(x)\neq 1$ for some $x\in\mathbb{Q}_{>0}$. Let $u\in\mathbb{Q}_{>0}$ be such that $g(u)\neq 1$ and that $\text{ht}\big(g(u)\big)$ is minimized. By (*), we see that $v:=g(u)$ satisfies $$\text{ht}\big(g(v)\big)=\text{ht}\big(g(u)\big)-1\,.$$
This contradicts the choice of $u$. Therefore, $u$ cannot exist, and the claim follows. This implies that
$$f(x)=\frac1x\text{ for all }x\in\mathbb{Q}_{>0}\,.$$
Best Answer
It might be helpful to standardize the functions and inputs to the functions. In other words, get rid of the $\exp$ and make things on the left and right look kind of similar:
$$ \begin{align*} &\exp \left( f (\log x )\right ) = \log \left( f(\exp x) \right) \\ \iff &f (\log x ) = \log \log \left( f(\exp x) \right) \\ \iff &f (\log \log x ) = \log \log f(x). \end{align*} $$
Let $g(x)$ be the function $g(x) = \log \log x$. Then the equality holds iff $f \circ g = g \circ f$.
Simple examples of $f$ that satisfy this is the identity $f(x) = x$ or the inverse of $g$ i.e. $f(x) = \exp \exp x$.
In a little more generality, $f$ can be the function that repeatedly exponentiates $k$ times, denoted $f(x) = \exp_k (x)$ since
$$ \begin{align*} \exp_k (\log \log x ) &= \exp_{k-2} (x) \\ &= \log \log \exp_k (x). \end{align*} $$
In fact you can allow $k$ to be $0$ or negative so that e.g. $\exp_0$ is the identity and $\exp_{-3}$ is $\log$ applied $3$ times, and the statement would still hold.
Structurally speaking $f = \exp_k$ works for any integer $k$ because $g(x) = \log \log x$ can be written as $g(x) = \exp_{-2}(x)$ and the question is asking for functions $f$ where $f \circ \exp_{-2} = \exp_{-2} \circ f$. The $\exp_k$ family of functions is closed under composition and also commutative since $\exp_a \circ \exp_b = \exp_b \circ \exp_a = \exp_{a + b}$.
I'm not sure if there is a more general characterization of functions that are commutative with $\log \log x$ but I can think about it.