How can you prove 3^n is not O(n^2)?
Let's assume that $3^n = O(n^2)$.

This would by definition mean that there are real positive numbers $a$ and $b$ such that for any sufficiently large $n$ (or to say as $lim_{n \to +\infty}$):

$ an^2 \leq 3^n \leq bn^2 $

Dividing everything by $3^n$ we get:

$ a\frac{n^2}{3^n} \leq 1 \leq b\frac{n^2}{3^n}$

And since if we let $lim_{n \to +\infty}$ we get a contradiction:

$ 0 \leq 1 \leq 0 $

Hence inital statement is false and $ 3^n \neq O(n^2) $