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

Now differentiating the above function we get,

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{\left(x^2-6x-16\right)·1-x·\left(2x-6\right)}{{\left(x^2-6x-16\right)}^2}$

$\Rightarrow f^{\text{'}}\left(x\right)=\frac{-x^2-16}{{\left(x^2-6x-16\right)}^2}$

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

Now, $x^2-6x-16\ne 0$

$\Rightarrow x^2-8x+2x-16\ne 0$

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

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

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

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