The solution of the differential equation ${\mathrm{xy}}^2 \mathrm{dy} – ({\mathrm{x}}^3+{\mathrm{y}}^3) \mathrm{dx} = 0$ is

$\begin{array}{rcl}{\mathrm{xy}}^2 \mathrm{dy}-\left({\mathrm{x}}^3+{\mathrm{y}}^3\right) \mathrm{dx}& =& 0\\ & \Rightarrow & {\mathrm{xy}}^2 \mathrm{dy}=\left({\mathrm{x}}^3+{\mathrm{y}}^3\right) \mathrm{dx}\\ & \Rightarrow & \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{x}}^3+{\mathrm{y}}^3}{{\mathrm{xy}}^2} -\left(1\right)\end{array}$

Now, Let $\begin{array}{rcl}\mathrm{y}& =& \mathrm{Vx} \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{V}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}} -\left(2\right)\\ \mathrm{V}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}& =& \frac{{\mathrm{x}}^3+{\mathrm{y}}^3}{{\mathrm{xy}}^2}=\frac{{\mathrm{x}}^3}{{\mathrm{xy}}^2}+\frac{{\mathrm{y}}^3}{{\mathrm{y}}^2\mathrm{x}}=\frac{{\mathrm{x}}^2}{{\mathrm{y}}^2}+\frac{\mathrm{y}}{\mathrm{x}}\\ \therefore \mathrm{V}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}& =& \frac{{\mathrm{x}}^2}{{\mathrm{y}}^2}+\frac{\mathrm{y}}{\mathrm{x}}\\ \mathrm{V}+\mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}& =& \frac{{\mathrm{x}}^2}{{\mathrm{y}}^2}+\mathrm{V}\\ & \Rightarrow & \mathrm{x}\frac{\mathrm{dv}}{\mathrm{dx}}=\frac{1}{{\displaystyle \raisebox{1ex}{${\mathrm{y}}^2$}\!\left/ \!\raisebox{-1ex}{${\mathrm{x}}^2$}\right.}}=\frac{1}{\sqrt{2}}\end{array}$

                      $\Rightarrow {\mathrm{V}}^2\mathrm{dv}=\frac{\mathrm{dx}}{\mathrm{x}} -\left(3\right)$

                      $\begin{array}{rcl}\int {\mathrm{V}}^2\mathrm{dv}& =& \int \frac{\mathrm{dx}}{\mathrm{x}}\\ & \Rightarrow & \frac{{\mathrm{V}}^3}{3^3}=\log \mathrm{x}+\log \mathrm{c}\end{array}$

                       $\begin{array}{rcl} & \Rightarrow & \frac{{\mathrm{V}}^3}{3}=\log \left(\mathrm{xc}\right)\\ & \Rightarrow & {\mathrm{V}}^3=3\log \left(\mathrm{cx}\right)\\ & \Rightarrow & \frac{{\mathrm{y}}^3}{{\mathrm{x}}^3}=3 \log \left(\mathrm{cx}\right)\\ & \Rightarrow & {\mathrm{y}}^3=3{\mathrm{x}}^3+\log \left(\mathrm{cx}\right)\end{array}$