${x^2}dy + (xy + {y^2})dx = 0;y = 1$ when $x = 1$

: ${x^2}dy + (xy + {y^2})dx = 0 \Rightarrow \cfrac{{dy}}{{dx}} = - \cfrac{{xy + {y^2}}}{{{x^2}}}$

$\Rightarrow \cfrac{{dy}}{{dx}} = - \left\{ {\cfrac{y}{x} + {{\left( {\cfrac{y}{x}} \right)}^2}} \right\}$,

which is a homogeneous differential equation.
To solve this equation,

we put $y = vx \Rightarrow \cfrac{{dy}}{{dx}} = v + x\cfrac{{dv}}{{dx}}$

Then, the differential equation becomes
$v + x\cfrac{{dv}}{{dx}} = - (v + {v^2}) \Rightarrow x\cfrac{{dv}}{{dx}} = - (2v + {v^2})$

$\Rightarrow$ $\cfrac{{dv}}{{2v + {v^2}}} = - \cfrac{{dx}}{x}$

…(2)
Integrating (2) both sides,

we get
$\cfrac{1}{2}\int {\left( {\cfrac{1}{v} - \cfrac{1}{{v + 2}}} \right)dv} = - \int {\cfrac{{dx}}{x}}$

$\Rightarrow \cfrac{1}{2}\log |v| - \cfrac{1}{2}\log |v + 2| = - \log |x| + \log |{C_1}|$

$\Rightarrow \cfrac{1}{2}\log \left| {\cfrac{v}{{v + 2}}} \right| = \log \left| {\cfrac{{{C_1}}}{x}} \right| \Rightarrow \cfrac{v}{{v + 2}} = {\left( {\cfrac{{{C_1}}}{x}} \right)^2}$

$\Rightarrow \cfrac{{\cfrac{y}{x}}}{{\cfrac{y}{x} + 2}} = \cfrac{{C_1^2}}{{{x^2}}} \Rightarrow \cfrac{y}{{y + 2x}} = \cfrac{{C_1^2}}{{{x^2}}}$

$\Rightarrow {x^2}y = C(2x + y)$ where $C_1^2 = C$

When $x = 1,y = 1$, then ${(1)^2}(1) = C(2 + 1) \Rightarrow C = \cfrac{1}{3}$
Putting value of $C$ in (3),

we get
${x^2}y = \cfrac{1}{3}(2x + y) \Rightarrow y + 2x = 3{x^2}y$

which is the required particular solution.