Factorise

$\frac{1}{3} x^{2}-2 x-9=\frac{x^{2}-6 x-27}{3}$

$=\frac{x^{2}-9 x+3 x-27}{3}$

$=\frac{x(x-9)+3(x-9)}{3}$

$=\frac{(x-9)(x+3)}{3}$

$=\frac{(x-9)}{3} \times \frac{(x+3)}{1}$

$=\left(\frac{1}{3} x-3\right)(x+3)$

Hence, factorisation of $\frac{1}{3} x^{2}-2 x-9$ is $\left(\frac{1}{3} x-3\right)(x+3)$.