Find $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ in terms of $x$ and $y$.

Determine the equation of the tangent to $C$ at the point $\left( {\frac{2}{{\text{e}}},{\text{ e}}} \right)$

$y + x\frac{{{\text{d}}y}}{{{\text{d}}x}} - \frac{1}{y}\frac{{{\text{d}}y}}{{{\text{d}}x}} = 0$     M1A1A1

Note:     Award A1 for the first two terms, A1 for the third term and the 0.

$\frac{{{\text{d}}y}}{{{\text{d}}x}} = \frac{{{y^2}}}{{1 - xy}}$     A1

Note:     Accept $\frac{{ - {y^2}}}{{\ln y}}$.

Note:     Accept $\frac{{ - y}}{{x - \frac{1}{y}}}$.

[4 marks]

${m_T} = \frac{{{{\text{e}}^2}}}{{1 - {\text{e}} \times \frac{2}{{\text{e}}}}}$     (M1)

${m_T} = - {{\text{e}}^2}$     (A1)

$y - {\text{e}} = - {{\text{e}}^2}x + 2{\text{e}}$

$- {{\text{e}}^2}x - y + 3{\text{e}} = 0$ or equivalent     A1

Note:     Accept $y = - 7.39x + 8.15$.

[3 marks]