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

\(\frac{d}{dx}(2x^3) = 6x^2\)

\(\frac{d}{dx}(-y^3) = -3y^2 \frac{dy}{dx}\)

\(\frac{d}{dx}(-3xy^2) = -3(y^2 + 2xy \frac{dy}{dx})\)

Combine and simplify:

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

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

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

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

(ii) For the tangent to be parallel to the y-axis, \(\frac{dx}{dy} = 0\), which implies \(\frac{dy}{dx}\) is undefined. This occurs when the denominator \(y^2 + 2xy = 0\).

Solve \(y^2 + 2xy = 0\):

\(y(y + 2x) = 0\)

\(y = 0\) or \(y = -2x\)

For \(y = 0\), substitute into the curve equation:

\(2x^3 = 2a^3\)

\(x = a\)

Point: \((a, 0)\)

For \(y = -2x\), substitute into the curve equation:

\(2x^3 - (-2x)^3 - 3x(-2x)^2 = 2a^3\)

\(2x^3 + 8x^3 - 12x^3 = 2a^3\)

-\(2x^3 = 2a^3\)

\(x = -a\)

\(y = -2(-a) = 2a\)

Point: \((-a, 2a)\)