This is my second response, after the clarifying remark from the asker (see below).

Let us assume that $m>e^e$ so that $\log\log m>0$. Then there is $m_0>m$ such that $f(m,n)<0$ for any $n\in(m,m_0)$. To see this claim, let $\epsilon>0$ be small, and consider $n$ such that
$$ \log n=(\log m)^{1+\epsilon}. $$ Then
$$ \frac{\log\log\log m}{\log\log\log n} \centerdot \frac{\log\log n}{\log\log m} = \frac{(1+\epsilon)\log\log\log m}{\log\log\log m+\log(1+\epsilon)} $$
$$=1+\epsilon-\frac{\epsilon}{\log\log\log m}+O(\epsilon^2)$$
while
$$ \frac{\log(\frac n m)}{(\log n)\log\log m}= \frac{1-(\log m)^{-\epsilon}}{\log\log m} = \frac{\epsilon\log\log m+O(\epsilon^2)}{\log\log m}=\epsilon+O(\epsilon^2).$$
Therefore
$$ f(m,n) = -\frac{\epsilon}{\log\log\log m}+O(\epsilon^2)$$
which is negative if $\epsilon>0$ is sufficiently small in terms of $m$.

**P.S.** My original response contained the more obvious claim that $f(m,n)>0$ when $n$ is sufficienly large in terms of $m$. Hence, for any $m>e^e$, there exist negative and positive values of $f(m,n)$ with $n>m$.