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