(i) Differentiate \(\ln(xy)\) implicitly: \(\frac{1}{xy}(y + x\frac{dy}{dx}) = \frac{1}{x} + \frac{1}{y}\frac{dy}{dx}\).

Differentiate \(y^3\): \(3y^2\frac{dy}{dx}\).

Set the derivatives equal: \(\frac{1}{x} + \frac{1}{y}\frac{dy}{dx} = 3y^2\frac{dy}{dx}\).

Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{y}{x(3y^3 - 1)}\).

(ii) The tangent is parallel to the y-axis when \(\frac{dy}{dx}\) is undefined, i.e., when \(3y^3 - 1 = 0\).

Solve \(3y^3 - 1 = 0\): \(y = 0.693\).

Substitute \(y = 0.693\) into the original equation: \(\ln(x \times 0.693) - (0.693)^3 = 1\).

Solve for \(x\): \(x = 5.47\).