The solution of $\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{y}}^2}{\mathrm{xy}-{\mathrm{x}}^2}$ is

$$\begin{gathered} Given equation can be written as \\ \frac{dx}{dy}=\frac{xy-x^2}{y^2} \\ \frac{dx}{dy}=\frac{\mathrm{x}}{\mathrm{y}}-{\left(\frac{\mathrm{x}}{\mathrm{y}}\right)}^2\to \left(1\right) \\ \mathrm{Take}\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{v} \\ \mathrm{diff} \mathrm{w}.\mathrm{r}.\mathrm{to} \mathrm{y} \mathrm{on} \mathrm{both} \mathrm{sides} \\ \frac{\mathrm{dx}}{\mathrm{dy}}=\mathrm{v}+\mathrm{y}\frac{\mathrm{dv}}{\mathrm{dy}} \\ \left(1\right)\Rightarrow \\ \mathrm{v}+\mathrm{y}\frac{\mathrm{dv}}{\mathrm{dy}}=\mathrm{v}-{\mathrm{v}}^2 \\ \mathrm{by} \mathrm{separating} \mathrm{variables} \\ \frac{-1}{{\mathrm{v}}^2}\mathrm{dv}=\frac{\mathrm{dy}}{\mathrm{y}} \\ \mathrm{integrating} \mathrm{on} \mathrm{both} \mathrm{sides} \\ \int \frac{-1}{{\mathrm{v}}^2}\mathrm{dv}=\int \frac{\mathrm{dy}}{\mathrm{y}} \\ \Rightarrow \frac{1}{\mathrm{v}}=\mathrm{logy}+\mathrm{logk} \\ \Rightarrow \frac{\mathrm{y}}{\mathrm{x}}=\mathrm{logky} \\ \Rightarrow \mathrm{ky}={\mathrm{e}}^{\frac{\mathrm{y}}{\mathrm{x}}} \end{gathered}$$