Answer
$\dfrac{f(a+h)-f(a)}{h}=\dfrac{-3}{a(a+h)}$
Work Step by Step
Using the result of item $47,$ $\dfrac{f(a+h)-f(a)}{h}=\dfrac{\dfrac{3}{a+h}-\dfrac{3}{a}}{h}
,$ then
\begin{array}{l}\require{cancel}
\dfrac{f(a+h)-f(a)}{h}=\dfrac{\dfrac{3}{a+h}\cdot\dfrac{a}{a}-\dfrac{3}{a}\cdot\dfrac{a+h}{a+h}}{h}
\\\\
\dfrac{f(a+h)-f(a)}{h}=\dfrac{\dfrac{3a}{a(a+h)}-\dfrac{3(a+h)}{a(a+h)}}{h}
\\\\
\dfrac{f(a+h)-f(a)}{h}=\dfrac{\dfrac{3a-3(a+h)}{a(a+h)}}{h}
\\\\
\dfrac{f(a+h)-f(a)}{h}=\dfrac{\dfrac{3a-3a-3h}{a(a+h)}}{h}
\\\\
\dfrac{f(a+h)-f(a)}{h}=\dfrac{\dfrac{-3h}{a(a+h)}}{h}
\\\\
\dfrac{f(a+h)-f(a)}{h}=\dfrac{-3h}{a(a+h)}\div h
\\\\
\dfrac{f(a+h)-f(a)}{h}=\dfrac{-3h}{a(a+h)}\cdot\dfrac{1}{h}
\\\\
\dfrac{f(a+h)-f(a)}{h}=\dfrac{-3\cancel{h}}{a(a+h)}\cdot\dfrac{1}{\cancel{h}}
\\\\
\dfrac{f(a+h)-f(a)}{h}=\dfrac{-3}{a(a+h)}
.\end{array}
