The folium of Descartes is a curve defined by the equation \({x^3} + {y^3} - 3xy = 0\), shown in the following diagram.

Determine the exact coordinates of the point P on the curve where the tangent line is parallel to the \(y\)-axis.

\({x^3} + {y^3} - 3xy = 0\)

\(3{x^2} + 3{y^2}\frac{{{\text{d}}y}}{{{\text{d}}x}} - 3x\frac{{{\text{d}}y}}{{{\text{d}}x}} - 3y = 0\)     M1A1

Note:     Differentiation wrt \(y\) is also acceptable.

\(\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{3y - 3{x^2}}}{{3{y^2} - 3x}}{\text{ }}\left( { = \frac{{y - {x^2}}}{{{y^2} - x}}} \right)\)     (A1)

Note:     All following marks may be awarded if the denominator is correct, but the numerator incorrect.

\({y^2} - x = 0\)     M1

EITHER

\(x = {y^2}\)

\({y^6} + {y^3} - 3{y^3} = 0\)     M1A1

\({y^6} - 2{y^3} = 0\)

\({y^3}({y^3} - 2) = 0\)

\((y \ne 0)\therefore y = \sqrt[3]{2}\)     A1

\(x = {\left( {\sqrt[3]{2}} \right)^2}{\text{ }}\left( { = \sqrt[3]{4}} \right)\)     A1

OR

\({x^3} + xy - 3xy = 0\)     M1

\(x({x^2} - 2y) = 0\)

\(x \ne 0 \Rightarrow y = \frac{{{x^2}}}{2}\)     A1

\({y^2} = \frac{{{x^4}}}{4}\)

\(x = \frac{{{x^4}}}{4}\)

\(x({x^3} - 4) = 0\)

\((x \ne 0)\therefore x = \sqrt[3]{4}\)     A1

\(y = \frac{{{{\left( {\sqrt[3]{4}} \right)}^2}}}{2} = \sqrt[3]{2}\)     A1

[8 marks]