Find $f\colon \Bbb{N}\to\Bbb{N}$ such that $f = n^{o(1)}$ and for every $c\in\Bbb{N}$, $f(n)=\omega(\log^c n)$
Take the log of your ratio to get
$$
c\log\log n - \frac{\log n}{\log\log n} = \frac{c\log^2 m - m}{\log m}, \quad \text{where } m = \log n.
$$
It is not hard to show that this tends to $-\infty$ as $m\to\infty$, and so as $n\to\infty$, which means that your ratio tends to zero as $n\to\infty$.

Indeed, since $c\log^2 m = o(m)$, for large enough $m$ we have $c \log^2 m - m \leq -m/2$. Since $2\log m = o(m)$, it follows that $\frac{m}{2\log m} \to \infty$ and so $\frac{-m}{2\log m} \to -\infty$.