Solving T(n) = 3T(n/3)+n/2 using master method
While applying master's theorem all you have to do is compare which of $f(n)$ or $(n^{\log_b{a}})$ is asymptotically greater. The greater of the two functions determines the solution to the recurrence. There are only possible outcomes for this comparison: 
[1] $(n^{\log_b{a}})$ is asymptotically greater than $f(n)$ 
[2] Both are asymptotically equal 
[3] $f(n)$ is asymptotically greater than $(n^{\log_b{a}})$ 

Yours falls in case 2.