I suppose you mean
[tex]f(x) = \dfrac{x+5}{x+1}[/tex]
Then
[tex]f(3) = \dfrac{3+5}{3+1} = \dfrac84 = 2[/tex]
and the difference quotient is
[tex]\dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5}{x+1}-2}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{\frac{x+5-2(x+1)}{x+1}}{x-3} \\\\ \dfrac{f(x)-f(3)}{x-3} = \dfrac{-x+3}{(x+1)(x-3)} \\\\ \dfrac{f(x)-f(3)}{x-3} = \boxed{-\dfrac{x-3}{(x+1)(x-3)}}[/tex]
If it's the case that x ≠ 3, then (x - 3)/(x - 3) reduces to 1, and you would be left with
[tex]\dfrac{f(x)-f(3)}{x-3}\bigg|_{x\neq3} = -\dfrac1{x+1}[/tex]
