I have showed that $f(n)=\Omega(g(n))$ in the following way.
We assume that $${\log n}^{\log n} \leq n/\log n$$ $$\implies \log n \times \log \log n \leq \log n - \log \log n$$ $$\implies \log \log n \times ( \log n -1 ) \leq \log n $$ which is a contradiction when $n \geq 1$. Thus we conclude $f(n) = \Omega(g(n))$ where we consider $c=1$ and $n_0 = 1$.
I will be thankful if anyone can comment that if my approach is correct or if incorrect may please provide the solution approach.