Answer
$- \dfrac{2(2x+h)}{((x+h)^2+16)(x^2+16)} \text{ or } \dfrac{-4x-2h}{((x+h)^2+16)(x^2+16)}$.
Work Step by Step
$\begin{array}{ l l }
\begin{array}{l}
=\dfrac{\dfrac{2\left( x^{2} +16\right)}{\left(( x+h)^{2} +16\right)\left( x^{2} +16\right)} -\dfrac{2\left(( x+h)^{2} +16\right)}{\left(( x+h)^{2} +16\right)( x^{2} +16}}{h}\\
\\
=\dfrac{\dfrac{2\left( x^{2} +16\right) -2\left(( x+h)^{2} +16\right)}{\left(( x+h)^{2} +16\right)\left( x^{2} +16\right)}}{h}\\
\end{array} & \begin{array}{l}
\mathrm{Apply\ the\ rule}\\
\dfrac{a}{b} -\dfrac{c}{d} =\dfrac{ad-bc}{bd}
\end{array}\\
& \\
\begin{array}{l}
=\dfrac{\dfrac{2x^{2} +32-2\left( x^{2} +2xh+h^{2} +16\right)}{\left(( x+h)^{2} +16\right)\left( x^{2} +16\right)}}{h}\\
\\
=\dfrac{\dfrac{2x^{2} +32-2x^{2} -4xh-2h^{2} -32}{\left(( x+h)^{2} +16\right)\left( x^{2} +16\right)}}{h}
\end{array} & \begin{array}{l}
\mathrm{Expand\ the\ expressions}\\
\mathrm{in\ the\ numerator\ by\ }\\
\mathrm{applying\ the\ distributive}\\
\mathrm{property}
\end{array}\\
& \\
=\dfrac{\dfrac{2x^{2} -2x^{2} +32-32-4xh-2h^{2}}{\left(( x+h)^{2} +16\right)\left( x^{2} +16\right)}}{h} & \mathrm{Group\ like\ terms\ together}\\
& \\
=\dfrac{\dfrac{-4xh-2h^{2}}{\left(( x+h)^{2} +16\right)\left( x^{2} +16\right)}}{h} & \mathrm{Simplify}\\
& \\
=\dfrac{\dfrac{-2h( 2x+h)}{\left(( x+h)^{2} +16\right)\left( x^{2} +16\right)}}{h} & \begin{array}{l}
\mathrm{Factor\ } -4xh-2h^{2} .\ \mathrm{The}\\
\mathrm{GCF\ is} \ -2h
\end{array}\\
& \\
=\dfrac{-h}{h} \cdot \dfrac{2( 2x+h)}{\left(( x+h)^{2} +16\right)\left( x^{2} +16\right)} & \mathrm{Factor\ out\ } -\dfrac{h}{h}\\
& \\
=-\dfrac{2( 2x+h)}{\left(( x+h)^{2} +16\right)\left( x^{2} +16\right)} & \mathrm{Simplify}
\end{array}$