For this you can use Akra-Bazzi method.
But recurrence graph can help equally well.
If you have $T(n)$ as top node, then in the previous step you have had these nodes (sum of which gives $T(n)$)
$$T(\frac{n}{3})|T(\frac{2n}{3})|n$$
The next level has had these nodes ($n$ is added from the previous level):
$$T(\frac{n}{9}),T(\frac{2n}{9}),\frac{n}{3}|T(\frac{2n}{9}),T(\frac{4n}{9}),\frac{2n}{3},n$$
So you have at this level
$$T(\frac{n}{9}),(2)T(\frac{2n}{9}),T(\frac{4n}{9}),2n$$
Notice that because $\frac{n}{3}+\frac{2n}{3}=n$ each level will have a node with $n$ as a value. (2) above signifies the number of nodes.
Instead of trying to do this manually one level at a time, the structure of nodes after $N$ steps is the same as what in symbolic notation you would have after deriving
$$(x\frac{1}{3}+y\frac{2}{3})^N=\sum_{k=0}^{N}\binom{N}{k}x^{N-k}\frac1{3^{N-k}}y^{k}\frac{2^k}{3^k}$$
Now you simply replace any $x^py^q$ with $T(n^*)$ and there you have the structure after $N$ steps, of course we will add $Nn$.
$$\sum_{k=0}^{N} \binom{N}{k}T(\frac{2^k}{3^N}n)+Nn$$
The question of how many levels you have is easily answered by looking at $\frac{2^k}{3^N}n$ because this one will never go lower than 1. So N is obviously $\sim \log(n)$ because $\frac{2^k}{3^N}$ is exponential.
So in the end we can safely assume $T(n) \sim n\log(n)$. Since we do not know what $T(1)$ is and this will affect the evaluation since this value will be multiplied by the number of the leaf nodes, we take $T(n) \cong an\log(n)$ and try to calculate the asymptotic value of $a$. Even though it is not necessary, we just want to confirm the finding.
$$an\log(n)=\frac{1}{3}an\log(\frac{n}{3})+\frac{2}{3}an\log(\frac{2n}{3})+n$$
$$\frac{1}{3}a\log(\frac{1}{3})+\frac{2}{3}a\log(\frac{2n}{3})+1=0$$
or
$$a=\frac{3}{3\log(3)+2\log(2)}$$
and there you have it:
$$T(n) \cong \frac{3}{3\log(3)+2\log(2)} n\log(n)$$
I assume by $f(n)$, you mean your recurrence relation to be of the form
$$
T(n)=aT\left(\frac{n}{b}\right)+f(n),
$$
so that in this case you have $a=3$, $b=2$, and $f(n)=0$ for all $n$.
That being the case: yes, $f(n)=0$ is perfectly acceptable.
To answer your more specific questions:
Yes, $0=O(n^{\log_2(3)})$. Take absolutely any $n_0\in\mathbb{N}$ and any $c>0$, and it is true that $0\leq cn^{\log_2(3)}$ for all $n\geq n_0$. Remember, Big-O just represents asymptotic upper bounds; of course something that goes to infinity is an upper bound, in the limit, for $0$.
Yes, $0$ is 'asymptotically smaller' than $n^{\log_2(3)}$. Take absolutely any $\epsilon\in(0, \log_2(3))$, and it is true that $0=O(n^{\epsilon})$.
The Master Theorem is perfectly applicable in this situation, and it shows that your $T(n)$ satisfies $T(n)=\Theta(n^{\log_2(3)})$.
Best Answer
No you can't ignore the -3, actually this case is a lot worse than the "normal" case.
For instance imagine $T(n)=2T(n-1)$, then it means $T(n)=\Theta(2^n)$, because you have to use the recurrence relation $n$ times instead of $log_b(n)$ times.
In your case it will be the same: you have $T(n)=\Theta(9^{n/3})=\Theta((\sqrt[3]9)^n)$