There are quite a number of comparisons to be made. It is in most cases relatively straightforward to decide about the relative long-term size.
Let's start with your pair. We have $f(n)=\log(n)/\log(\log (n))$ and $g(n)=n^{\log(n)}$.
For large $n$ (and it doesn't have to be very large), we have $f(n)\lt \log(n)$.
Also, for large $n$, we have $\log(n)\gt 1$, and therefore $g(n)=n^{\log(n)}\gt n$.
So for large $n$, we have $\frac{f(n)}{g(n)} \lt \frac{\log(n)}{n}.$ But we know that $\lim_{n\to\infty}\dfrac{\log(n)}{n}=0$. This can be shown in various ways. For instance, we can consider $\dfrac{\log(x)}{x}$ and use L'Hospital's Rule to show this has limit $0$ as $x\to\infty$.
It takes less time to deal with the pair $n/\log(n)$ and $n^{\log(n)}$. If $n$ is even modestly large, we have $n/\log(n)\lt n$. But after a while, $\log(n)\gt 2$, so $n^{\log(n)}\gt n^2$. It follows that in the long run, $n^{\log(n)}$ grows much faster than $n/\log(n)$.
As a last example, let us compare $\log^3(n)$ and $n/\log(n)$. Informally, $\log$ grows glacially slowly. More formally, look at the ratio $\dfrac{\log^3(n)}{n/\log(n)}$. This simplifies to $\frac{\log^4 (n)}{n}.$ We can use L'Hospital's Rule on $\log^4(x)/x$. Unfortunately we then need to use it several times. It is easier to examine $\dfrac{\log(x)}{x^{1/4}}$. Then a single application of L'Hospital's Rule does it. Or else we can let $x=e^y$. Then we are looking at $y^4/e^y$, and we can quote the standard result that the exponential function, in the long run, grows faster than any polynomial.
Remark: The second person in your list is the slowest one. Apart from that, they are in order. So you only need to prove four facts to get them lined up. A fair part of the work has been done above.