Answer
$y=(-e-1)x+1$
Work Step by Step
$xe^y+ye^x=1$
$(0,1)$
$e^y+xe^y\frac{dy}{dx}+\frac{dy}{dx}e^x+ye^x=0$
$e^y+ye^x+\frac{dy}{dx}(xe^y+e^x)=0$
$\frac{dy}{dx}=\frac{-(e^y+ye^x)}{xe^y+e^x}$
$\frac{dy}{dx}(0)=-(e+1)$
Equation of Tangent:
$y-1=-(e+1)(x-0)$
$y-1=(-e-1)x$
$y=(-e-1)x+1$