Answer
$\displaystyle \frac{dy}{dx}=\frac{6}{(2-3x )^{2}}$
Work Step by Step
$f(x)=3x,\displaystyle \ \ \quad g(x)=2-3x,\quad y=\frac{f(x)}{g(x)}$
$f^{\prime}(x)=3\qquad g^{\prime}(x)=-3$
$\displaystyle \frac{dy}{dx}=\frac{d}{dx}[\frac{f(x)}{g(x)}]$= ... quotient rule ...
$= \displaystyle \frac{f^{\prime}(x)g(x)-f(x)g^{\prime}(x)}{[g(x)]^{2}}$
$=\displaystyle \frac{3(2-3x ) - 3x(-3)}{(2-3x )^{2}}$
$=\displaystyle \frac{6-9x+9x}{(2-3x )^{2}}$
$= \displaystyle \frac{6}{(2-3x )^{2}}$