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 \((- 3, 0)\).