$xy = \log y + C:y' = \cfrac{{{y^2}}}{{1 - xy}}(xy \ne 1)$

.: We have, $xy = \log y + C$
…(1)
Differentiating (1) w.r.t. $x$,

we get
$xy' + y = \cfrac{1}{y}y'$

$\Rightarrow xy'y + {y^2} = y'y \Rightarrow {y^2} = y' - xy'y$

$\Rightarrow {y^2} = y'(1 - xy) \Rightarrow y' = \cfrac{{{y^2}}}{{1 - xy}}$

$\therefore$ $xy = \log y + C$ is a solution of
the given differential equation.