General Rules of Thumb which hold for sufficiently large $n$:
$$\begin{align}
&\log^c(n) \leq n^{\epsilon}, \quad \text{ for any } \epsilon > 0, \quad \text { and any constant } c
\\
\\
&n^k \leq k^n, \qquad \text{where } k \text{ is any constant } k > 1
\end{align}
$$
As Chris noted above, we can write this as
$$
A \ll \log^c (n) \ll n^{\epsilon} \ll k^n
$$
To actually prove these rules, you just compute limits (assuming you know calculus). To see that $\log (n) \ll n^{\varepsilon}$ for example, just note that
$$
\lim_{n \to \infty} \frac{\log n}{n^{\epsilon}} = \lim_{n \to \infty} \frac{n^{-1}}{\epsilon n^{\epsilon - 1}} = 0 = \lim_{n \to \infty} \frac{1}{\epsilon n^{\epsilon}} = 0.
$$
where $A$ and $c$ are constants, $\epsilon > 0$, and $k$ is a real number such that $k > 1$.
If you want to show that $\log^c(n) \ll n^{\epsilon}$ for any positive integer $c$, then just note that
$$
\lim_{n \to \infty} \frac{\log^c (n)}{n^\epsilon} \leq \left( \lim_{n \to \infty} \frac{\log (n)}{n^{\epsilon/c}} \right) \dots \left( \lim_{n \to \infty} \frac{\log(n)}{n^{\epsilon/ c}}\right) = 0
$$
by what we showed above. We can actually take $c$ be to be any arbitrary real number and the above still holds. This is easy to see since every real number is less than or equal to some integer (infinitely many in fact!). I will leave it to you to show that $n \log^3 (n) \ll n^{4/3}$, for example. The more complicated functions work the same way but we might need to be slightly creative when evaluating the limits.
Hope this helps.
You could just compare the two functions. The base 3 is constant and the base $n$ tends to infinity. Now compare the exponents: $\sqrt{\log n}$ grows more slowly than $\log n$. These two observations imply that $3^{\sqrt{\log n}}$ grows more slowly that $n^{\log n}$.
In fact, for $n>10$:
$$
{ n^{\log n}\over3^{\sqrt{\log n}}}\ge { 10^{\log n}\over3^{\sqrt{\log n}}}
\ge{ 10^{\log n}\over3^{ \log n }}={(10/3)^{\log n}}\rightarrow\infty.
$$
Best Answer
To answer @sol4me If you know some better way then please share, i will be glad to know.
First, never trust a plot. You just saw it may hurt, so even if it can help, it's not a proof.
Then, you must know some basic comparison scales, for example, as $n\rightarrow \infty$, and fixed $a>0, b>0, c>1$ and any real $d$,
$$d \ll \log \log n \ll \log^a n \ll n^b \ll c^n \ll n! \ll n^n$$
Where I write $a_n \ll b_n$ iff $\frac{a_n}{b_n} \rightarrow 0$ (this is not a standard notation !)
Of course, there are many other possible asymptotic comparisons, these are just the most frequent.
You have also some allowed operations, for example,
You prove such things by computing the limit. Taking $\log$ as you did may be very useful (for example in the first case above).
Finally, you have to apply these comparisons to your case, and sometimes it's a bit tricky, but honestly here it's not.
When you have a doubt, write what the comparison means as a limit, and compute the limit.
And remember, these comparisons are asymptotic. Sometimes the smallest $n$ such that the inequality really holds may be very large, but it's nevertheless only a fixed, finite number. You may have inequality for $n> 10^{10^{10}}$ for example, so trying to plot things is often hopeless.
If you want to know more about such methods, there are good readings, such as Concrete Mathematics, by Graham, Knuth and Patashnik.