(i) Differentiate the equation \(x^3 + 2y^3 = 3xy\) implicitly with respect to \(x\):

\(3x^2 + 6y^2 \frac{dy}{dx} = 3y + 3x \frac{dy}{dx}\).

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

\(6y^2 \frac{dy}{dx} - 3x \frac{dy}{dx} = 3y - 3x^2\).

\(\frac{dy}{dx} (6y^2 - 3x) = 3y - 3x^2\).

\(\frac{dy}{dx} = \frac{3y - 3x^2}{6y^2 - 3x} = \frac{y - x^2}{2y^2 - x}\).

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

Set \(y - x^2 = 0\) which gives \(y = x^2\).

Substitute \(y = x^2\) into the original equation:

\(x^3 + 2(x^2)^3 = 3x(x^2)\).

\(x^3 + 2x^6 = 3x^3\).

\(2x^6 = 2x^3\).

\(x^3(x^3 - 1) = 0\).

\(x^3 = 0\) or \(x^3 = 1\).

\(x = 0\) or \(x = 1\).

For \(x = 1\), \(y = 1^2 = 1\).

Thus, the point is \((1, 1)\).