Explanation

(a)
We have, $\left( xy - x^{2} \right) = y^{2}$
$\Rightarrow$ $y^{2}\ \ \frac{dx}{dy} = xy - x^{2}$
$\Rightarrow$ $\frac{1}{x^{2}}\ \ \ \frac{dx}{dy} - \frac{1}{x}.\frac{1}{y} = - \frac{1}{y^{2}}$
Put $\frac{1}{x} = v\ \ \Rightarrow - \frac{1}{x^{2}}\ \ \frac{dx}{dy} = \frac{dv}{dy}$
$\therefore$ $\frac{dv}{dy} + \frac{v}{y} = \frac{1}{y^{2}}$, which is linear
$\therefore$ $IF = e^{\int_{}^{}{\frac{1}{y}\ \ dy}} = e^{\log y} = y$
$\therefore$ The solution is $vy = \int_{}^{}\frac{1}{y^{2}}\ y\ dy + c$
$\Rightarrow$ $\frac{y}{x} = \log y + c$
$\Rightarrow$ $y = x\ (\log y + c)$
This passes through the point (-1,1)
$\therefore$ $1 = - 1(\log 1 + c)$
ie., $c = - 1$
thus, the equation of the curve is
$y = x(\log y - 1)$