(i) Differentiate both sides of \(\sin y \ln x = x - 2 \sin y\) with respect to \(x\).

Using the product rule on the left side, we have:

\(\frac{d}{dx}(\sin y \ln x) = \cos y \frac{dy}{dx} \ln x + \sin y \frac{1}{x}\).

The derivative of the right side is:

\(1 - 2\cos y \frac{dy}{dx}\).

Equating the derivatives, we get:

\(\cos y \frac{dy}{dx} \ln x + \frac{\sin y}{x} = 1 - 2\cos y \frac{dy}{dx}\).

Rearrange to solve for \(\frac{dy}{dx}\):

\(\frac{dy}{dx}(\cos y \ln x + 2\cos y) = 1 - \frac{\sin y}{x}\).

\(\frac{dy}{dx} = \frac{1 - \frac{\sin y}{x}}{\cos y \ln x + 2\cos y}\).

(ii) For the tangent to be parallel to the \(x\)-axis, \(\frac{dy}{dx} = 0\).

Set the numerator of \(\frac{dy}{dx}\) to zero:

\(1 - \frac{\sin y}{x} = 0\).

\(x = \sin y\).

Substitute \(x = \sin y\) into the original equation:

\(\sin y \ln \sin y = \sin y - 2\sin y\).

\(\ln \sin y = -1\).

\(\sin y = \frac{1}{e}\).

Thus, \(x = \frac{1}{e}\).