(i) Differentiate \(y = \frac{1 + e^{-x}}{1 - e^{-x}}\) using the quotient rule:

\(\frac{dy}{dx} = \frac{(1 - e^{-x})(-e^{-x}) - (1 + e^{-x})(-e^{-x})}{(1 - e^{-x})^2}\)

Simplify to \(\frac{dy}{dx} = \frac{-2e^{-x}}{(1 - e^{-x})^2}\).

Since \(e^{-x} > 0\) for \(x > 0\), \(-2e^{-x} < 0\). Therefore, \(\frac{dy}{dx}\) is always negative.

(ii) Set \(\frac{dy}{dx} = -1\):

\(\frac{-2e^{-a}}{(1 - e^{-a})^2} = -1\)

Simplify to \(2e^{-a} = (1 - e^{-a})^2\)

Let \(u = e^{-a}\), then \(2u = (1 - u)^2\)

Expand and simplify: \(2u = 1 - 2u + u^2\)

Rearrange to \(u^2 - 4u + 1 = 0\)

Solving this quadratic equation gives \(u = 2 \pm \sqrt{3}\)

Since \(u = e^{-a} > 0\), choose \(u = 2 - \sqrt{3}\)

Thus, \(e^{-a} = 2 - \sqrt{3}\)

\(a = -\ln(2 - \sqrt{3}) = \ln(2 + \sqrt{3})\)