$x y^{2} d y-\left(x^{3}+y^{3}\right) d x=0$
$\Rightarrow xy ^{2} dy =\left( x ^{3}+ y ^{3}\right) d x$ (changing sides)
$\Rightarrow \frac{d y}{d x}=\frac{x^{3}+y^{3}}{x y^{2}}$ ...(i)
Now, Let $y = vx$
$\Rightarrow \frac{ dy }{ dx }= v + x \frac{ dv }{ dx }$ ..(ii)
replacing value of $\frac{d y}{d x}$ in (i)
$v + x \frac{ dv }{ dx }=\frac{ x ^{3}+ y ^{3}}{ xy ^{2}}$
$=\frac{ x ^{3}}{ xy ^{2}}+\frac{ y ^{3}}{ y ^{2} x }=\frac{ x ^{2}}{ y ^{2}}+\frac{ y }{ x }$
$\therefore v + x \frac{ dv }{ dx }=\frac{ x ^{2}}{ y ^{2}}+\frac{ y }{ x }$
$v + x \frac{ dv }{ dx }=\frac{ x ^{2}}{ y ^{2}}+ v$
$\Rightarrow x \frac{ dv }{ dx }=\frac{1}{ y ^{2} / x ^{2}}=\frac{1}{ v ^{2}}$
$\Rightarrow v ^{2} dv =\frac{ d x }{ x }$ ...(iii)
(by cross multiplication)
Now, integrating both side of equation (iii)
we have,
$\int v ^{2} dv =\int \frac{ dx }{ x }$
$\Rightarrow \frac{v^{3}}{3}=\log x+\log c$
$\Rightarrow \frac{v^{3}}{3}=\log (x c)$
$[\because \log a+\log b=\log a b]$
$\Rightarrow v^{3}=3 \log (c x)$
$\Rightarrow \frac{y^{3}}{x^{3}}=3 \log (c x)$
$y^{3}=3 x^{3} \log (c x)$