The inverse function of $$f(x) = \frac{8^{2x} - 8^{-2x}}{8^{2x} + 8^{-2x}}$$, $$x \in (-1, 1)$$, is

Let us begin by writing the given relation in a form that can be inverted. We put

$$y \;=\; f(x)\;=\;\frac{8^{2x}-8^{-2x}}{8^{2x}+8^{-2x}},\qquad x\in(-1,1).$$

To clear the negative exponent we multiply the numerator and the denominator by $$8^{2x}$$. This gives

$$y \;=\;\frac{8^{2x}\!\cdot\!8^{2x}\;-\;1}{8^{2x}\!\cdot\!8^{2x}\;+\;1}\;=\;\frac{8^{4x}-1}{8^{4x}+1}.$$

Now we eliminate the fraction by cross-multiplication:

$$y\bigl(8^{4x}+1\bigr)=8^{4x}-1.$$

Expanding the left side we obtain

$$y\cdot8^{4x}+y \;=\;8^{4x}-1.$$

We bring all terms containing $$8^{4x}$$ to one side and the constants to the other side:

$$y\cdot8^{4x}-8^{4x} \;=\;-1-y.$$

Factoring out $$8^{4x}$$ on the left gives

$$8^{4x}(y-1)\;=\;-(1+y).$$

We divide by $$y-1$$ (remembering that $$y\neq1$$ inside the interval of definition) and, at the same time, remove the minus sign from numerator and denominator:

$$8^{4x}\;=\;\frac{-(1+y)}{\,y-1\,}\;=\;\frac{1+y}{1-y}.$$

The base-exponent form $$a^m=b$$ can always be rewritten as the logarithmic form $$m=\log_a b$$. Applying this rule with base $$8$$ we take logarithm base $$8$$ of both sides:

$$4x\;=\;\log_8\!\left(\frac{1+y}{1-y}\right).$$

Dividing by $$4$$ produces

$$x\;=\;\frac14\,\log_8\!\left(\frac{1+y}{1-y}\right).$$

Up to this point $$y$$ was the original output and $$x$$ was the original input. For the inverse function we interchange their roles, writing the final answer as

$$f^{-1}(x)\;=\;\frac14\,\log_8\!\left(\frac{1+x}{1-x}\right).$$

This expression exactly matches Option D.

Hence, the correct answer is Option D.