Answer
$\dfrac{x^2+9}{-6x}$
Work Step by Step
The given expression, $
\dfrac{\dfrac{x-3}{x+3}+\dfrac{x+3}{x-3}}{\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{\dfrac{(x-3)^2+(x+3)^2}{(x+3)(x-3)}}{\dfrac{(x-3)^2-(x+3)^2}{(x+3)(x-3)}}
\\\\=
\dfrac{\dfrac{(x-3)^2+(x+3)^2}{\cancel{(x+3)(x-3)}}}{\dfrac{(x-3)^2-(x+3)^2}{\cancel{(x+3)(x-3)}}}
\\\\=
\dfrac{(x-3)^2+(x+3)^2}{(x-3)^2-(x+3)^2}
\\\\=
\dfrac{(x^2-6x+9)+(x^2+6x+9)}{(x^2-6x+9)-(x^2+6x+9)}
\\\\=
\dfrac{x^2-6x+9+x^2+6x+9}{x^2-6x+9-x^2-6x-9}
\\\\=
\dfrac{(x^2+x^2)+(-6x+6x)+(9+9)}{(x^2-x^2)+(-6x-6x)+(9-9}
\\\\=
\dfrac{2x^2+18}{-12x}
\\\\=
\dfrac{2(x^2+9)}{-12x}
\\\\=
\dfrac{\cancel{2}(x^2+9)}{\cancel{2}\cdot(-6)x}
\\\\=
\dfrac{x^2+9}{-6x}
.\end{array}