Two cowboys A and B decide to solve a dispute with a duel.

Let $p$ be the probability that A wins. With probability ${1\over3}$ he wins on his first shot. With probability ${2\over3}$ he misses. When B hits in his first shot, A has lost. When B misses, A again has a chance of $p$. It follows that $$p={1\over3}+{2\over3}\cdot{1\over3}\cdot p\ .$$ This implies $p={3\over7}$.

What we have is a geometric series starting from $\frac13$ and the next term $\frac29$ of the previous. The sum is $$\frac13\cdot\frac1{1-2/9}=\frac37$$

You are definitely on the right track. A hint might be that A wins at turn k ($A_k$) iff all the previous shots fail (both A's and B's), and the last one does not (and they are all mutually exclusive events).
The chance of both of them failing is ($\frac{1}{3}\frac{2}{3} = \frac{2}{9}$), and the chance of A hitting the target is 1/3. Therefore the expression you look for is: $$P(\text{A wins}) = P(\cup_{k\in\mathbb{N}}A_k) = \sum_{k=0}^{\infty} (\frac{2}{9})^k (1/3)$$