@himgta big oh 'O' is used for asymptotic upper bounds not asymptotic tight upper bounds. We use small 'o' for functions which are not asymptotic tight upper bounds.
So , here I have to show $g(n) = n^3$is not asymptotic tight upper bound.
So, to show something is not asymptotic upper bound , we have to use definition of small 'o' notation.
Definition of $f(n) = o(g(n))$ says $f(n) \leq cg(n)$ for all $c>0$ and $n \geq n_{0}$ . Here difference b/w big O and small o is that small o is used for all $c>0$ and big O is used for some $c>0$
So, here we have to know 3n log(n!) + (n^2+3)log n $ \leq cn^3$ is true or not for all $c>0$ or not.
In other way we can write $f(n) = o(g(n))$ as $\lim_{n\rightarrow \infty }\frac{f(n)}{g(n)} = 0$
So, overall if we prove $\lim_{n\rightarrow \infty }\frac{3nlogn! + n^{2}logn +3logn }{n^{3}} = 0$ then we say that it is not a tight upper bound.
here, for n! , use stirling's approximation and use L-Hospitals's rule here, then it will give 0 , so it will be not be tight upper bound. So, remaining function $n^2logn$ will be tight upper bound.These things are given in cormen. So, you can verify it.