Answer
$\dfrac{x-3}{3}$
Work Step by Step
The given expression, $
\dfrac{\dfrac{x}{3}-\dfrac{3}{x}}{1+\dfrac{3}{x}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{\dfrac{x^2-9}{3x}}{\dfrac{x+3}{x}}
\\\\=
\dfrac{x^2-9}{3x}\div\dfrac{x+3}{x}
\\\\=
\dfrac{x^2-9}{3x}\cdot\dfrac{x}{x+3}
\\\\=
\dfrac{(x+3)(x-3)}{3x}\cdot\dfrac{x}{x+3}
\\\\=
\dfrac{(\cancel{x+3})(x-3)}{3\cancel{x}}\cdot\dfrac{\cancel{x}}{\cancel{x+3}}
\\\\=
\dfrac{x-3}{3}
.\end{array}
