The function $$f(x) = \frac{x}{x^2 - 6x - 16}, x \in \mathbb{R} - \{-2, 8\}$$

Consider the function $$f(x) = \frac{x}{x^2 - 6x - 16}$$ whose domain is $$\mathbb{R} - \{-2, 8\}$$. To determine its monotonicity, we compute its derivative using the quotient rule:

$$ f'(x) = \frac{(x^2 - 6x - 16)\cdot 1 - x\cdot (2x - 6)}{(x^2 - 6x - 16)^2}. $$

The numerator simplifies as follows:

$$ x^2 - 6x - 16 - 2x^2 + 6x = -x^2 - 16 = -(x^2 + 16). $$

Therefore,

$$ f'(x) = \frac{-(x^2 + 16)}{(x^2 - 6x - 16)^2}. $$

Since $$x^2 + 16 > 0$$ for all real $$x$$, the numerator is always negative, and the denominator $$(x^2 - 6x - 16)^2$$ is always positive (being a perfect square and nonzero in the domain). Hence $$f'(x) < 0$$ on every interval of its domain.

It follows that $$f(x)$$ is decreasing on $$(-\infty, -2)\cup(-2, 8)\cup(8, \infty)$$.

The correct answer is Option (2).