A curve has equation ${x^3}{y^2} + {x^3} - {y^3} + 9y = 0$. Find the coordinates of the three points on the curve where $\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$.

$3{x^2}{y^2} + 2{x^3}y\frac{{{\text{d}}y}}{{{\text{d}}x}} + 3{x^2} - 3{y^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} + 9\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$     M1M1A1

Note:     First M1 for attempt at implicit differentiation, second M1 for use of product rule.

$\left( {\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3{x^2}{y^2} + 3{x^2}}}{{3{y^2} - 2{x^3}y - 9}}} \right)$

$\Rightarrow 3{x^2} + 3{x^2}{y^2} = 0$     (A1)

$\Rightarrow 3{x^2}\left( {1 + {y^2}} \right) = 0$

$x = 0$     A1

Note:     Do not award A1 if extra solutions given eg $y =  \pm 1$.

substituting $x = 0$ into original equation     (M1)

${y^3} - 9y = 0$

$y(y + 3)(y - 3) = 0$

$y = 0,{\text{ }}y =  \pm 3$

coordinates $(0, 0), (0, 3), (0, - 3)$     A1

[7 marks]

The majority of candidates were able to apply implicit differentiation and the product rule correctly to obtain $3{x^2}\left( {1 + {y^2}} \right) = 0$. The better then recognised that $x = 0$ was the only possible solution. Such candidates usually went on to obtain full marks. A number decided that $y =  \pm 1$ though then made no further progress. The solution set $x = 0$ and $y =  \pm i$ was also occasionally seen. A small minority found the correct x and y values for the three co-ordinates but then surprisingly expressed them as ${\text{(0, 0), (3, 0)}}$ and $(- 3, 0)$.