Answer
\[y=\ln (e^x+e-e^3)\]
Work Step by Step
Slope of tangent to a curve at ($x,y$) is given by $\frac{dy}{dx}$ at ($x,y$)
\[\frac{dy}{dx}=e^{x-y}=\frac{e^x}{e^y}\]
Seperating variables
\[e^y dy=e^x dx\]
Integrating
\[\int e^y dy=\int e^x dx+C\]
Where C is constant of integration
$e^y=e^x+C$ _____(1)
Curve passes through (3,1)
\[e=e^3+C
\Rightarrow C=e-e^3\]
From (1)
\[e^y=e^x+e-e^3\]
\[y=\ln (e^x+e-e^3)\]
Hence equation of curve is $y=\ln (e^x+e-e^3)$
