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

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

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

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

Set the derivative equal to zero:

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

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

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

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

(ii) To find the equation of the tangent at \((2, 1)\), first find the gradient:

Substitute \(x = 2\) and \(y = 1\) into \(\frac{dy}{dx} = \frac{3x^2 - 2xy}{x^2 + 3y^2}\):

\(\frac{dy}{dx} = \frac{3(2)^2 - 2(2)(1)}{(2)^2 + 3(1)^2} = \frac{12 - 4}{4 + 3} = \frac{8}{7}\)

The equation of the tangent line is:

\(y - 1 = \frac{8}{7}(x - 2)\)

Rearrange to the form \(ax + by + c = 0\):

\(7(y - 1) = 8(x - 2)\)

\(7y - 7 = 8x - 16\)

\(8x - 7y - 9 = 0\)