[Cambridge/9709/2009]
The equation of a curve is $x^3 - x^2y - y^3 = 3$.

1. Differentiate the equation implicitly with respect to $x$, applying the product rule to the $x^2y$ term.

$\frac{\text{d}}{\text{d}x}(x^3) - \frac{\text{d}}{\text{d}x}(x^2y) - \frac{\text{d}}{\text{d}x}(y^3) = \frac{\text{d}}{\text{d}x}(3)$
$3x^2 - \left(2xy + x^2 \frac{\text{d}y}{\text{d}x}\right) - 3y^2 \frac{\text{d}y}{\text{d}x} = 0$

2. Group the terms containing $\frac{\text{d}y}{\text{d}x}$ and factorise.

$3x^2 - 2xy - x^2 \frac{\text{d}y}{\text{d}x} - 3y^2 \frac{\text{d}y}{\text{d}x} = 0$
$3x^2 - 2xy = x^2 \frac{\text{d}y}{\text{d}x} + 3y^2 \frac{\text{d}y}{\text{d}x}$
$3x^2 - 2xy = \frac{\text{d}y}{\text{d}x}(x^2 + 3y^2)$

3. Make $\frac{\text{d}y}{\text{d}x}$ the subject of the formula.

$\frac{\text{d}y}{\text{d}x} = \frac{3x^2 - 2xy}{x^2 + 3y^2}$

1. Obtain the expression for $\frac{\text{d}y}{\text{d}x}$ from part (i).

By implicitly differentiating $x^3 - x^2y - y^3 = 3$ with respect to $x$:
$3x^2 - \left(x^2\frac{\text{d}y}{\text{d}x} + 2xy\right) - 3y^2\frac{\text{d}y}{\text{d}x} = 0$
$\frac{\text{d}y}{\text{d}x}(x^2 + 3y^2) = 3x^2 - 2xy$
$\frac{\text{d}y}{\text{d}x} = \frac{3x^2 - 2xy}{x^2 + 3y^2}$

2. Evaluate the gradient of the tangent at the point $(2, 1)$.

Substitute $x = 2$ and $y = 1$ into the derivative expression:
$m = \frac{3(2)^2 - 2(2)(1)}{(2)^2 + 3(1)^2}$
$m = \frac{12 - 4}{4 + 3} = \frac{8}{7}$

3. Formulate the equation of the tangent and rearrange it into the required form.

Using the point-gradient formula $y - y_1 = m(x - x_1)$:
$y - 1 = \frac{8}{7}(x - 2)$
$7(y - 1) = 8(x - 2)$
$7y - 7 = 8x - 16$
$8x - 7y - 9 = 0$