The function f(x) = x x ^ 2 - 6x - 16 , x ∈ ℝ - \ - 2, 8\

Given: $f\left(x\right) = \dfrac{x}{{x}^{2} - 6x - 16}$

Now differentiating the above function we get,

$⇒ {f}^{'}\left(x\right) = \dfrac{\left({x}^{2} - 6x - 16\right) \cdot 1 - x \cdot \left(2x - 6\right)}{{\left({x}^{2} - 6x - 16\right)}^{2}}$

$⇒ {f}^{'}\left(x\right) = \dfrac{- {x}^{2} - 16}{{\left({x}^{2} - 6x - 16\right)}^{2}}$

$⇒ {f}^{'}\left(x\right) < 0 . . . \left(i\right)$

Now, ${x}^{2} - 6x - 16 \ne 0$

$⇒ {x}^{2} - 8x + 2x - 16 \ne 0$

$⇒ \left(x + 2\right)\left(x - 8\right) \ne 0$

$⇒ x \ne 8, - 2 . . . \left(ii\right)$

Using \left(i\right)\text{and}\left(ii\right),

So,$the function is always decreasing inx \in ( - \infty , - 2 ) \cup ( - 2, 8 ) \cup ( 8, \infty )$.