Answer
y=1
Work Step by Step
$1+\ln xy=e^{x-y}$
$0+\frac{1}{x}y+\ln(x)\frac{dy}{dx}=(1-\frac{dy}{dx})e^{x-y}$
$0+\frac{1}{x}y+\ln(x)\frac{dy}{dx}=e^{x-y}-e^{x-y}\frac{dy}{dx}$
$(e^{x-y}+\ln x)\frac{dy}{dx}=e^{x-y}-\frac{y}{x}$
$\frac{dy}{dx}=\frac{e^{x-y}-\frac{y}{x}}{(e^{x-y}+\ln x)}$
at point (1,1),
$\frac{dy}{dx}=\frac{e^{1-1}-\frac{1}{1}}{(e^{1-1}+\ln1)}$
$=\frac{e^0-1}{e^0+0}$
$=0$
Equation of tangent:
y-1=0(x-1)
y-1=0
y=1
