Solve this

We are given with an equation $e^{x-y}=\log \left(\frac{x}{y}\right)=\log x-\log y$, we have to find $\frac{d y}{d x}$ of it, so by differentiating the equation on both sides with respect to $x$, we get,

$e^{x-y}\left(1-\frac{d y}{d x}\right)=\frac{1}{x \ln 10}-\frac{1}{y \ln 10} \frac{d y}{d x}$

$\frac{d y}{d x}\left[\frac{1}{y \ln 10}-e^{x-y}\right]=\frac{1}{x \ln 10}-e^{x-y}$

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\frac{1}{\mathrm{x} \ln 10}-\mathrm{e}^{\mathrm{x}-\mathrm{y}}}{\frac{1}{\mathrm{y} \ln 10}-\mathrm{e}^{\mathrm{x}-\mathrm{y}}}$

$\frac{d y}{d x}=\frac{\frac{1-x \ln 10 e^{x-y}}{x}}{\frac{1-y \ln 10 e^{x-y}}{y}}=\frac{y\left(1-x \ln 10 e^{x-y}\right)}{x\left(1-y \ln 10 e^{x-y}\right)}$