The $(r+1)$ th term in the expansion of ${\left[ {(3/2){x^2} – (1/3x)} \right]^9}$ is given by

${T_{r + 1}} = {\,^9}{C_r}{\left( {\frac{3}{2}{x^2}} \right)^{9 – r}}{\left( { – \frac{1}{{3x}}} \right)^r} = {\,^9}{C_r}{( – 1)^r}\frac{{{3^{9 – 2r}}}}{{{2^{9 – r}}}}{x^{18 – 3r}}$      …..$(1)$

Since we are looking for the coefficient of the term independent of $x$ in the expansion of

$\left(1+x+2 x^{3}\right)\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$         …..$(2)$

we must get the coefficient of $x^{0}, x^{-1}$ and $x^{-3}$ in the expansion of ${\left[ {(3/2){x^2} – (1/3x)} \right]^9}.$

For $x^{0}, r$ must be $6$ in $( 1);$ for $x^{-1},$ there is no value of $r ;$ and for $x^{-3}, r$ must be $7$

in $(1) .$ Therefore, the coefficient of the term independent of $x$ in $( 2)$ is

$1.{\,^9}{{\rm{C}}_6}{( – 1)^6} \cdot \frac{{{3^{9 – 12}}}}{{{2^{9 – 6}}}} + 2 \cdot {\,^9}{{\rm{C}}_7}{( – 1)^7} \cdot \frac{{{3^{9 – 14}}}}{{{2^{9 – 7}}}}$

${=\frac{9.8 .7}{1.2 .3} \cdot \frac{3^{-3}}{2^{3}}+2 \frac{9.8}{1.2}(-1) \cdot \frac{3^{-5}}{2^{2}}=\frac{7}{18}-\frac{2}{27}=\frac{17}{54}}$