Answer
$\dfrac{x}{2-3x}$
Work Step by Step
The given expression, $
\dfrac{\dfrac{2}{x}+3}{\dfrac{4}{x^2}-9}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{\dfrac{2+x(3)}{x}}{\dfrac{4-x^2(9)}{x^2}}
\\\\=
\dfrac{\dfrac{2+3x}{x}}{\dfrac{4-9x^2}{x^2}}
\\\\=
\dfrac{2+3x}{x}\div\dfrac{4-9x^2}{x^2}
\\\\=
\dfrac{2+3x}{x}\cdot\dfrac{x^2}{4-9x^2}
\\\\=
\dfrac{2+3x}{x}\cdot\dfrac{x\cdot x}{(2-3x)(2+3x)}
\\\\=
\dfrac{\cancel{2+3x}}{\cancel{x}}\cdot\dfrac{x\cdot \cancel{x}}{(2-3x)(\cancel{2+3x})}
\\\\=
\dfrac{x}{2-3x}
.\end{array}
