Let $f \prec g$ denote "f grows asymptotically slower than g", then you can use the following easy rule for polylogarithmic? functions:
$$n^{\alpha_1}(\log n)^{\alpha_2}(\log \log n)^{\alpha_3} \prec n^{\beta_1}(\log n)^{\beta_2}(\log \log n)^{\beta_3} \Longleftrightarrow (\alpha_1, \alpha_2, \alpha_3) < (\beta_1, \beta_2, \beta_3)$$
The order relation between the tuples is lexicographic. I.e. $(2, 10) < (3, 5)$ and $(2, 10) > (2, 5)$
Applied to your example:
$\mathcal{O}(n/\log n) \Rightarrow (1, -1, 0)$
$\mathcal{O}(n^{2/3}) \Rightarrow (2/3, 0, 0)$
$\mathcal{O}(n^{1/3}) \Rightarrow (1/3, 0, 0)$
$$(1/3, 0, 0) < (2/3, 0, 0) < (1, -1, 0)\\
\Rightarrow \mathcal{O}(n^{1/3}) \prec \mathcal{O}(n^{2/3}) \prec \mathcal{O}(n/\log n)$$
You could say: powers of n dominate powers of log, which dominate powers of log log.
Source: Concrete Mathematics, p. 441