Find of function.

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

Taking logarithm on both the sides, we obtain

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

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

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

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

Differentiating both sides with respect to x, we obtain

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

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

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

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

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