Let $$f(x) = \frac{1}{7 - \sin 5x}$$ be a function defined on $$\mathbb{R}$$. Then the range of the function $$f(x)$$ is equal to :

To begin, we find the range of the function $$f(x) = \frac{1}{7 - \sin 5x}$$.

The sine function satisfies $$-1 \leq \sin\theta \leq 1$$ for all real $$\theta$$, and since $$5x$$ spans all real values as $$x$$ ranges over $$\mathbb{R}$$, it follows that

$$-1 \leq \sin 5x \leq 1$$

The extreme values occur when $$\sin 5x = 1$$ (i.e.\ $$5x = \frac{\pi}{2} + 2k\pi$$) and $$\sin 5x = -1$$ (i.e.\ $$5x = -\frac{\pi}{2} + 2k\pi$$).

Next, we consider the denominator $$7 - \sin 5x$$. Multiplying the inequality $$-1 \leq \sin 5x \leq 1$$ by $$-1$$ reverses it, giving

$$-1 \leq -\sin 5x \leq 1$$

and adding 7 yields

$$6 \leq 7 - \sin 5x \leq 8$$

Since $$7 - \sin 5x$$ lies in $$[6,8]$$ and is always positive, we can take reciprocals. Because the function $$g(t) = \frac{1}{t}$$ is strictly decreasing for $$t > 0$$, reversing the inequality gives

$$\frac{1}{8} \leq \frac{1}{7 - \sin 5x} \leq \frac{1}{6}$$

We verify that these bounds are attained: when $$\sin 5x = -1$$, one finds $$f(x) = \frac{1}{7 - (-1)} = \frac{1}{8}$$, and when $$\sin 5x = 1$$, $$f(x) = \frac{1}{7 - 1} = \frac{1}{6}$$.

Therefore, the range of $$f(x)$$ is $$\left[\frac{1}{8}, \frac{1}{6}\right]$$. The correct answer is Option D: $$\left[\frac{1}{8}, \frac{1}{6}\right]$$.