Solve this

The given function is $x y=e^{x-y}$

Taking log on both sides, we obtain

$\log (x y)=\log \left(e^{x-y}\right)$

$\log x+\log y=(x-y) \log e$

$\log x+\log y=(x-y) \times 1$

$\log x+\log y=x-y$

Differentiating both sides with respect to $x$, we obtain

$\frac{\mathrm{d}}{\mathrm{dx}}(\log x)+\frac{\mathrm{d}}{\mathrm{dx}}(\log y)=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x})-\frac{\mathrm{dy}}{\mathrm{dx}}$

$\frac{1}{x}+\frac{1}{y} \frac{d y}{d x}=1-\frac{d y}{d x}$

$\left(1+\frac{1}{y}\right) \frac{d y}{d x}=1-\frac{1}{x}$

$\left(\frac{y+1}{y}\right) \frac{d y}{d x}=\frac{x-1}{x}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}(\mathrm{x}-1)}{\mathrm{x}(\mathrm{y}+1)}$