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

\(\frac{d}{dx}(x^3) - \frac{d}{dx}(3x^2y) + \frac{d}{dx}(y^3) = 0\).

This gives \(3x^2 - (6xy + 3x^2 \frac{dy}{dx}) + 3y^2 \frac{dy}{dx} = 0\).

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

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

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

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

Set the numerator of \(\frac{dy}{dx} = \frac{x^2 - 2xy}{x^2 - y^2}\) to zero:

\(x^2 - 2xy = 0\).

Factor to find \(x(x - 2y) = 0\), giving \(x = 0\) or \(x = 2y\).

Substitute \(x = 0\) into the original equation:

\(y^3 = 3\), so \(y = 1.44\).

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

\((2y)^3 - 3(2y)^2y + y^3 = 3\).

\(8y^3 - 12y^3 + y^3 = 3\).

\(-3y^3 = 3\), so \(y = -1\) and \(x = -2\).

The points are \((-2, -1)\) and \((0, 1.44)\).