Does $f(n) + g(n) = O(g(n) \cdot f(n))$ hold? I would like to know if this statement is true:
$f(n) + g(n) = O(g(n)\cdot f(n))$.

I thought of giving a counter example by defining:
$f(n) = 3n^2$
; $g(n) = n$ which will give us that $O(3n^3) = n^3$ but i'm not sure if it's possible to say that $O(3n^3) = 3n^2 + n$ because I suspect that it might be right because of the definition of Big O. And if that's right that means that the counter example i gave is not good.

Is this a valid approach?
I don't get what you do with your choice of $f$ and $g$ but they don't work as a counter-example.

Here is how you see that:

$\qquad f(n) + g(n) = 3n^2 + n \in \Theta(n^2) \subseteq O(n^3) = O(f \cdot g)$.

See our reference question on how to show this.

It is easy to see that the same happens for all pairs of increasing functions: 

$\qquad f+g \in \Theta(\max(f,g)) \subseteq O(f \cdot g)$.

So if you want to have a chance at disproving the claim, you'll have to look for decreasing functions -- if your definition of Landau symbols admits such.