The term independent of $x$  in the expansion of ${\left(x^2-\frac{1}{3x}\right)}^9$  is equal to

Given expansion ${\left(x^2-\frac{1}{3x}\right)}^9$

We have general term in the expansion ${\left(x+a\right)}^n$

$(\therefore T_{r+1}={}^n{\text{C}}_r{x}^{n-r}{\left(a\right)}^r$ be the expansion of ${(x+a)}^n)$  

Then it can be written in the form of

$T_{r+1}={}^9{C}_r{\left(x^2\right)}^{9-r}{\left(-\frac{1}{3x}\right)}^r$

$T_{r+1}={}^9{C}_r{\left(x\right)}^{18-2r}{(-1)}^r{\left(x\right)}^{-r}{\left(\frac{1}{3}\right)}^r$

$={}^9{C}_r\frac{x^{18-3r}{(-1)}^r}{3^r}\mathrm{..............}\left(1\right)$  

The term independent of $x$  in the expansion means compare

$x^{18-3r}$  with $x^0$

This will not contain$x$ if  $x^{18-3r}=x^0$

$18-3r=0$

i.e. if $r=6$

$\therefore$  The term independent of $x$ substitute in Eq (1)

$\begin{array}{c}={}^9{C}_6\frac{{\left(-1\right)}^6}{3^6}({\therefore }^n{C}_r=\frac{n!}{(n-r)!r!})\\ =\frac{9\times 8\times 7}{1\times 2\times 3\times 3^6}=\frac{28}{243}\end{array}$