Given function is $\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{15}$

We know, the standard formula of $T_{r+1}$ will be,

$T_{r+1}={ }^{15} C_{r}\left(\frac{3 x^{2}}{2}\right)^{15-r}\left(-\frac{1}{3 x}\right)^{r}$

$T_{r+1}={ }^{15} C_{r}(-1)^{r} 3^{15-2 r} 2^{r-15} x^{30-3 r}$

The term independent of $x$ is,

$30-3 r=0$

Which implies $r=10$

On putting the value of $r$ in above obtained expression we get

$T_{10+1}={ }^{15} C_{10} 3^{-5} 2^{-5}$

$={ }^{15} C_{10}\left(\frac{1}{6}\right)^{5}$