Math Mixed Problem

I can see where you got your answer of $1\frac{7}{9}$.  $3\frac{1}{3}$ is not $3\times\frac{1}{3}$, $3\frac{1}{3}$ is $3+\frac{1}{3}$. 

$\frac{2}{3}\times3\frac{1}{3}+\frac{1}{3}(3+\frac{1}{3})$

$\frac{2}{3}\times(3+\frac{1}{3})+\frac{1}{3}(3+\frac{1}{3})$

$\frac{2}{3}(3+\frac{1}{3})+\frac{1}{3}(3+\frac{1}{3})$

$(\frac{6}{3}+\frac{2}{9})+\frac{1}{3}(3+\frac{1}{3})$

$(\frac{18}{9}+\frac{2}{9})+\frac{1}{3}(3+\frac{1}{3})$

$\frac{20}{9}+\frac{1}{3}(3+\frac{1}{3})$

$\frac{20}{9}+\frac{3}{3}+\frac{1}{9}$

$\frac{20}{9}+\frac{9}{9}+\frac{1}{9}$

$\frac{29}{9}+\frac{1}{9}$

$\frac{30}{9}$

$3\frac{3}{9}$

$3\frac{1}{3}$