$ydx + (x - {y^2})dy = 0$

: The given equation is $ydx + (x - {y^2})dy = 0$

$\Rightarrow \cfrac{{dx}}{{dy}} + \cfrac{x}{y} = y$

…(1)
which is a linear equation of the type $\cfrac{{dx}}{{dy}} + Px = Q$

Here, $P = \cfrac{1}{y}$ and $Q = y$

$\therefore$ ${\rm{I}}{\rm{.F}}. = {e^{\int {\cfrac{1}{y}dy} }} = {e^{\log \left| y \right|}} = y$

$\therefore$ The solution is $x(I.{\rm{F}}.) = \int Q ({\rm{I}}.{\rm{F}}.)dy + C$

$\Rightarrow xy = \int {y \cdot ydx} + C \Rightarrow xy = \int {{y^2}dy} + C$

$\Rightarrow xy = \cfrac{1}{3}{y^3} + C \Rightarrow x = \cfrac{{{y^2}}}{3} + \cfrac{C}{y}$,

which is the required solution