A calculus problem by Akshat Sharda

If we put $z = \frac{x(y-x)}{y} = x - \frac{x^2}{y}$ then $\frac{dz}{dx} \; = \; 1 - \frac{2x}{y} + \frac{x^2}{y^2}\,\frac{dy}{dx}$ so the differential equation becomes $\begin{aligned} e^z & = \; \frac{2x}{y} - 2 - \frac{x^2}{y^2}\frac{dy}{dx} \; = \; -\frac{dz}{dx} - 1 \\ \frac{dx}{dz} & = \; -\frac{1}{e^z+1} \; = \; - \frac{e^{-z}}{e^{-z}+1} \\ x + c & = \; \ln\big(e^{-z}+1\big) \\ Ce^x & = \; e^{-z}+1 \\ -z & = \; \ln(Ce^x - 1) \\ x(x-y) & = \; y\ln(Ce^x-1) \end{aligned}$ putting $C = e^c$ .