When $\left|x\right|<\frac{1}{2},$ the coefficient of $x^4$ in the expansion of $\frac{3x^2-5x+3}{(x-1)(2x+1)(x+3)}$ is

Rewrite the given expression

$F=-\left(x^2-\frac{5}{3}x+1\right)(1-x{)}^{-1}(1+2x{)}^{-1}{\left(1+\frac{x}{3}\right)}^{-1}$

$=\left\{\begin{array}{l}-\left(x^2-\frac{5}{3}x+1\right)\left(1+x+x^2+x^3+x^4\right)\\ \left(1-2x+4x^2-8x^3+16x^4\right)\left(1-\frac{x}{3}+\frac{x^2}{9}-\frac{x^3}{27}+\frac{x^4}{81}\right)\end{array}\right\}$

On neglecting higher degree terms

$=-\left(1+x+x^2+x^3+x^4-\left(\frac{5}{3}x+\frac{5}{3}{x}^2+\frac{5}{3}{x}^3+\frac{5}{3}{x}^4\right)+x^2+x^3+x^4\right)$

Simplify the above expression.

$=-\left(1-\frac{2}{3}x+\frac{x^2}{3}+\frac{x^3}{3}+\frac{x^4}{3}\right)\left(1-\frac{7x}{3}+\frac{43}{9}{x}^2-\frac{259}{27}{x}^3+\frac{1555}{81}{x}^4\right)$

$=-\frac{2166}{81}$

$=\frac{-722}{27}$