Answer would be D)
recurrence would be T(N) = T($\frac{N}{3}$) + T($\frac{2N}{3}$) + N
solve it using tree so $\frac{2^kN}{3^k}$= 1 at at K+1 level so,k=$log_\frac{3}{2}$n
at each level workdone is n so n+n+n+.....(k+1)time ,
$n*(k+1)$=$n$*($log_\frac{3}{2}$n+1)=$\theta($n$log_\frac{3}{2}$n)