If $x^y=e^{x-y}$, then $\frac{dy}{dx}=$

$\frac{logx}{(1+logx)^2}$

$x^y=e^{x-y}$

Take $log_e$​ on both sidees

$log_ex^y=log_ee^{x-y}⇒ ylogx=(x−y)log_e​e $

$ylogx=x−y$ ....(i)

$y[logx+1]=x$ ....(ii)

Differentiate both sides w.r.t. x

$\frac{dy}{dx}[1+logx]y[\frac{1}{x}]=1$

$\frac{dy}{dx}=\frac{1-\frac{y}{x}}{1+logx}$

from eq. (ii) $\frac{y}{x}=\frac{1}{(1+logx)}$

$\frac{dy}{dx}=\frac{1-\frac{1}{1+logx}}{1+logx}=\frac{logx}{(1+logx)^2}$