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