$f(n) = 3n^3 - 39n^2 + 360n + 20$. We need constants $c, n_0 > 0$ such that $f(n) \geq$ c$n^3$ $\,$ $\forall n\geq n_0$. Fix $c=2.25$. The smallest integer value that works can be found using calculus to find the range where the functions have $0$'s. The factors are equal to (n - 39.9762)(n - 12.0791)(n + 0.055248). Upon further inspection that for $-0.055248 \leq n \leq 12.0791$ and $n\geq 39.9762$ we have that $f(n) \geq 2.25n^3$. In order to satisfy $\forall n\geq n_0$ $f(n) \geq 2.25n^3$ we can't take $n_0$ in the range -0.55248 to 12.0791 since, for instance $n=13$ (greater than any of those numbers) would not satisfy the inequality. Thus, we can set $n_0 = 40$, the smallest integer value that satisfies are constraints.
My question is that you could also solve this by a sort of guess and check, but you would need to know the derivative of $f(n) - cg(n)$ (note that $g(n)$ would be equal to $n^3$ in this case) to see when the function stops decreasing in order to know that your $n_0$ is correct, and so how would you do it?
I think your approach to this problem is too complicated. It is easy to see that $f(n)-cn^3\ge 0$ when $n\ge 0$ and $(3-c)n^3\ge 39 n^2$, so for each $0<c<3$ we can put $n_0=\tfrac {39}{3-c}$. I guess that to find the smallest integer $n_0$ such that $f(n)\ge cn^3$ for each $n\ge n_0$ usually we have to find the largest root of the cubic equation $f(n)-cn^3=0$. This is a rater complicated task, but it can be done analytically, using Cardano formula.
In a general case, given $f(n)=a_dn^d+a_{d-1}n^{d-1}+\dots+ a_1n+a_0$ with $a_d>0$, for each $n\ge 1$ we have $f(n)\ge a_dn^d-Mn^{d-1}$, where $M=\sum_{i=0}^{d-1} |a_i|$, so for each $0<c<a_d$ we can we have $f(n)\ge cn^{d}$ for each $n\ge \tfrac M{a_d-c}$.