(i) To find \(\frac{dy}{dx}\), use the product rule: \(\frac{d}{dx}(uv) = u'v + uv'\).

Let \(u = e^{-2x}\) and \(v = \tan x\).

Then \(u' = -2e^{-2x}\) and \(v' = \sec^2 x\).

So, \(\frac{dy}{dx} = (-2e^{-2x}) \tan x + e^{-2x} \sec^2 x\).

Factor out \(e^{-2x}\): \(\frac{dy}{dx} = e^{-2x}(-2 \tan x + \sec^2 x)\).

Rewrite as \(e^{-2x}(-2 + \sec^2 x)\) using \(\sec^2 x = 1 + \tan^2 x\).

Thus, \(\frac{dy}{dx} = e^{-2x}(-2 + 1 + \tan^2 x) = e^{-2x}(a + b \tan x)^2\) where \(a = 1\) and \(b = 1\).

(ii) The gradient \(\frac{dy}{dx} = e^{-2x}(-2 + \sec^2 x)\) is never negative because \(\sec^2 x \geq 1\), making \(-2 + \sec^2 x \geq -1\).

(iii) The gradient is least when \(\sec^2 x\) is minimized, which occurs when \(x = \frac{1}{4} \pi\).