解:
$\begin{aligned}\frac{3x}{4x - 3}\div\frac{2}{16x^2 - 9}\cdot\frac{x}{4x + 3}&=\frac{3x}{4x - 3}\cdot\frac{(4x + 3)(4x - 3)}{2}\cdot\frac{x}{4x + 3}\\&=\frac{3x\cdot(4x + 3)(4x - 3)\cdot x}{(4x - 3)\cdot2\cdot(4x + 3)}\\&=\frac{3}{2}x^2\end{aligned}$