Ta có \(f\left( {{x_1}} \right) - f\left( {{x_2}} \right) = \left( {\dfrac{{{x_1} - 3}}{{{x_1} + 5}}} \right) - \left( {\dfrac{{{x_2} - 3}}{{{x_2} + 5}}} \right)\)

\( = \dfrac{{\left( {{x_1} - 3} \right)\left( {{x_2} + 5} \right) - \left( {{x_2} - 3} \right)\left( {{x_1} + 5} \right)}}{{\left( {{x_1} + 5} \right)\left( {{x_2} + 5} \right)}} = \dfrac{{8\left( {{x_1} - {x_2}} \right)}}{{\left( {{x_1} + 5} \right)\left( {{x_2} + 5} \right)}}\).

Với mọi \({x_1},{\rm{ }}{x_2} \in \left( { - \infty ; - 5} \right)\) và \({x_1} < {x_2}\). Ta có \(\left\{ \begin{array}{l}{x_1} < - 5\\{x_2} < - 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_1} + 5 < 0\\{x_2} + 5 < 0\end{array} \right.\).

Suy ra \(\dfrac{{f\left( {{x_1}} \right) - f\left( {{x_2}} \right)}}{{{x_1} - {x_2}}} = \dfrac{8}{{\left( {{x_1} + 5} \right)\left( {{x_2} + 5} \right)}} > 0 \Rightarrow f\left( x \right)\) đồng biến trên \(\left( { - \infty ; - 5} \right)\).

Với mọi \({x_1},{\rm{ }}{x_2} \in \left( { - 5; + \infty } \right)\) và \({x_1} < {x_2}\). Ta có \(\left\{ \begin{array}{l}{x_1} > - 5\\{x_2} > - 5\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}{x_1} + 5 > 0\\{x_2} + 5 > 0\end{array} \right.\).

Suy ra \(\dfrac{{f\left( {{x_1}} \right) - f\left( {{x_2}} \right)}}{{{x_1} - {x_2}}} = \dfrac{8}{{\left( {{x_1} + 5} \right)\left( {{x_2} + 5} \right)}} > 0 \Rightarrow f\left( x \right)\) đồng biến trên \(\left( { - 5; + \infty } \right)\).