Adapt a one-tape turing machine's algorithm that find the center of a string in O(nlog(n)) to find the first third
Let's start with a new way of computing the center. First of all, it's not so clear what to do with a string of even length. Fortunately, it doesn't matter, since we can fix it once and for all: initially, we determine the parity of the length of the input, and then add one symbol at the beginning or end (according to the definition of center for strings of even length).
From now on, we assume that we are given a string of odd length. Here is the idea in a nutshell:
\begin{align}
&amp;\verb|XXXXX| &amp;\verb|XXXXXXX| \\
&amp;\verb|X X X| &amp;\verb| X X X |
\end{align}
In words, we erase part of the input so that the remaining string has the same center. How we do this exactly depends on whether the input has length $4n+1$ or $4n+3$.

Let's move on to your case. The $1/3-2/3$ midpoint is well-defined for strings of length $3m+1$. You can reduce to this case by adding one or two symbols to either end of the string as before, so let us assume that our string has length $3m+1$. We again consider two cases, according to the parity of $m$:
\begin{align}
&amp;\verb|XXCXXXX| &amp;\verb|XXXCXXXXXX| \\
&amp;\verb|X C X X| &amp;\verb| X C X X Y|          
\end{align}
We have highlighted the center using the letter C.
If the string has length $6n+1$, then we simply cross of every other letter. If the string has length $6n+4$, then we cross of every other letter (starting from the second), and then remove the final letter (which we denoted by Y in the example).