Let $\frac{1}{\left(1-2x\right)^{2} \left(1-3x\right)} = \frac{A}{1-2x}+ \frac{B}{\left(1-2x\right)^{2}} + \frac{C}{1-3x} $
$ \therefore \, 1 =A\left(1-2x\right)\left(1-3x\right) +B\left(1-3x\right)+ C \left(1-2x\right)^{2}$ ...(i)
Putting $ x = \frac{1}{3}$ in (i), we get
$ 1 =0+0+C \left(1-\frac{2}{3}\right)^{2} \Rightarrow 1 = \frac{1}{9} C \Rightarrow C = 9 $
Putting $ x =\frac{1}{2}$ in (i), we get
$ \therefore \, 1 =B\left(1- \frac{3}{2}\right)\Rightarrow - \frac{1}{2}B = 1 \Rightarrow B = - 2$
Putting $ x =0$ in (i), we get $ 1=A+B+C$
$ A = 1-B-C = 1+2 - 9 = -6$
$ \therefore \, \frac{1}{\left(1-2x\right)^{2}\left(1-3x\right)} = \frac{-6}{1-2x} - \frac{2}{\left(1-2x\right)^{2}} + \frac{9}{1-3x}$
One of the terms is $ \frac{-6}{1-2x}$