How to solve recursion T(n) = T(n/3) + T(2n/3) + n?
Look at the recursion from QuickSort for the average case:
$T(n) = T(n/2) + T(n/2) + n$
which ends up in $O (n \log n)$.
Note that $n/2 + n/2$ sums up exactly to $n$. Just like $n/3 + 2n/3$ sums up exactly to $n$. Substitute the formula for $T(n)$ in $T(n/3)$ and $T(2n/3)$ and you'll see it's the sum of $T(x)$ for four values $x$ which add up to $n$, plus $2n$. Substitute again, and you see it's the sum of $T(x)$ for eight smaller values $x$ which add up to $n$, plus $3n$. And so on. Since the $x$ shrink by at least a factor $2/3$, the number of rounds until you end up at $T(1)$ is $O(\log n)$.
So that's what you guess, and once you guessed it you can prove it: $T(n) = O (n\log n)$.