(i) Let \(u = 1 + 3 \cos^2 x\). Then \(y = \frac{2}{3} \ln(u)\).

Using the chain rule, \(\frac{dy}{dx} = \frac{2}{3} \cdot \frac{1}{u} \cdot \frac{du}{dx}\).

We have \(\frac{du}{dx} = -6 \cos x \sin x = -3 \sin(2x)\).

Thus, \(\frac{dy}{dx} = \frac{2}{3} \cdot \frac{-3 \sin(2x)}{1 + 3 \cos^2 x}\).

Using \(\sin(2x) = \frac{2 \tan x}{1 + \tan^2 x}\) and \(\cos^2 x = \frac{1}{1 + \tan^2 x}\), we get:

\(\frac{dy}{dx} = \frac{-4 \tan x}{4 + \tan^2 x}\).

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

\(\frac{-4 \tan x}{4 + \tan^2 x} = -1\).

Solving, \(4 \tan x = 4 + \tan^2 x\).

\(\tan^2 x - 4 \tan x + 4 = 0\).

Solving the quadratic equation, \(\tan x = 2 \pm \sqrt{0} = 2\).

Thus, \(x = \arctan(2) \approx 1.11\).