How to compare rate of growth for following functions?
In other words, is $f(n)$ = $O( \, g(n) \, )$ or $g(n)$ = $O(\, f(n) \, )$?
$f(n) = n^{\frac{4}{3}}$ and $g(n) = n*(\log(n))^3$
How to solve these kind of questions in general (Basic methodologies on how to approach these questions) ? Are limits always bound to give right answer?
Yes.
By definition, $$f=O(g) \iff \frac{|f|}{|g|} \text {is bounded}$$
In your case, this translates to considering $$\lim _{n \to \infty} \frac{n^{4/3}}{n(\log n)^3} = \lim _{n \to \infty} \frac {n^{1/3}} {(\log n)^3}$$
Now use L'Hopital's rule three times:
\begin{align} \lim \frac {n^{1/3}} {(\log n)^3} & = \lim \frac{\frac{1}{3}n^{-2/3}} {3 (\log n)^2 (1/n)} \\ & = \lim \frac{n^{1/3}} {9 (\log n)^2} \\ &= \lim \frac{\frac{1}{3}n^{-2/3}}{18 (\log n) (1/n)} \\ &= \lim \frac{n^{1/3}}{54 \log n} \\ &= \lim \frac{\frac{1}{3}n^{-2/3}}{54 (1 / n)} \\ &= \frac{n^{1/3}}{162} \end{align}
which is clearly unbounded i.e. $f$ is not in $O(g)$.
Performing the converse analysis however, we find that $\lim _{n \to \infty} g/f = 0$, and therefore $g = O(f)$.