We induct on $n$. For $n=1$, the claim is vacuously true because there are no directed graphs on $1$ vertex with this property; that can be our base case.
Choose a vertex $v$ with out-degree at least $1$ (such a vertex must exist, because all vertices have in-degree at least $1$) and construct the breadth-first-search tree $T$ of all vertices that can be reached from $v$.
First, I claim that we can find a dominating set for $T$ of size at most $\frac23 |T|$. If $|T|=2$, then $\{v\}$ is a dominating set for $T$ that works. If $|T|=3$, there are two sets that are definitely dominating sets of $T$:
- $\{v\}$, together with all vertices reached in an even number of steps from $v$.
- $\{v\}$, together with all vertices reached in an odd number of steps from $v$.
The sum of their sizes is $|T|+1$, since both include $v$ and split the other vertices between them, so one of them has size at most $\frac{|T|+1}{2} \le \frac23 |T|$.
Second, I claim that if we remove $T$ from the graph $G$, then every vertex of the remaining graph still has in-degree at least $1$. If $w$ is a vertex of $G$ not in $T$, and $x$ is some vertex of $G$ with an edge $(x,w)$, then $x$ is not in $T$ - or else we'd have added $w$ to $T$ as well. So every vertex of $G-T$ still has an edge into it from another vertex of $G-T$.
By the inductive hypothesis, $G-T$ has a dominating set of size at most $\frac23(n - |T|)$. Together with the dominating set for $T$ of size at most $\frac23|T|$, we end up getting a dominating set of size at most $\frac23 n$ for all of $G$.