[Cambridge/9709/2006]
The equation of a curve is $x^3 + 2y^3 = 3xy$.

1. Differentiate the left-hand side of the equation with respect to $x$.

$\frac{\text{d}}{\text{d}x}(x^3 + 2y^3) = 3x^2 + 6y^2 \frac{\text{d}y}{\text{d}x}$

2. Differentiate the right-hand side using the product rule.

$\frac{\text{d}}{\text{d}x}(3xy) = 3y + 3x \frac{\text{d}y}{\text{d}x}$

3. Equate the derivatives and collect terms involving $\frac{\text{d}y}{\text{d}x}$.

$3x^2 + 6y^2 \frac{\text{d}y}{\text{d}x} = 3y + 3x \frac{\text{d}y}{\text{d}x}$

$6y^2 \frac{\text{d}y}{\text{d}x} - 3x \frac{\text{d}y}{\text{d}x} = 3y - 3x^2$

4. Factorise $\frac{\text{d}y}{\text{d}x}$ and simplify to arrive at the given expression.

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

$(2y^2 - x)\frac{\text{d}y}{\text{d}x} = y - x^2$

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

1. Apply the condition for a tangent parallel to the $x$-axis to find a relationship between $x$ and $y$.

A tangent parallel to the $x$-axis means $\frac{\text{d}y}{\text{d}x} = 0$.
$\frac{y - x^2}{2y^2 - x} = 0$
$y - x^2 = 0 \implies y = x^2$

2. Substitute $y = x^2$ into the equation of the curve.

$x^3 + 2(x^2)^3 = 3x(x^2)$
$x^3 + 2x^6 = 3x^3$

3. Solve the resulting polynomial equation for $x$.

$2x^6 - 2x^3 = 0$
$2x^3(x^3 - 1) = 0$
$x^3 = 0 \quad \text{or} \quad x^3 - 1 = 0$
$x = 0 \quad \text{or} \quad x = 1$

4. Reject the origin, find the required $y$-coordinate, and check validity.

The question specifies the point is 'other than the origin', so we reject $x = 0$.
For $x = 1$:
$y = (1)^2 = 1$
(Check that the denominator $2y^2 - x = 2(1)^2 - 1 = 1 \neq 0$, so the derivative is defined.)

The coordinates of the point are $(1, 1)$.