Answer
$\dfrac{1}{(a+1)(a+h+1)}$
Work Step by Step
$f(x)=\dfrac{x}{x+1}$
Find $f(a)$ by substituting $x$ with $a$:
$f(a)=\dfrac{a}{a+1}$
Find $f(a+h)$ by substituting $x$ with $a+h$:
$f(a+h)=\dfrac{a+h}{a+h+1}$
We have $f(a)$ and $f(a+h)$, substitute them in the expression $\dfrac{f(a+h)-f(a)}{h}$ and simplify:
$\dfrac{f(a+h)-f(a)}{h}=\dfrac{\dfrac{a+h}{a+h+1}-\dfrac{a}{a+1}}{h}=...$
Evaluate the substraction indicated in the numerator and simplify:
$...=\dfrac{\dfrac{(a+h)(a+1)-a(a+h+1)}{(a+1)(a+h+1)}}{h}=...$
$...=\dfrac{\dfrac{a^{2}+a+ah+h-a^{2}-ah-a}{(a+1)(a+h+1)}}{h}=\dfrac{\dfrac{h}{(a+1)(a+h+1)}}{h}=...$
$...=\dfrac{1}{(a+1)(a+h+1)}$