The correct answer is A:\(e^x+e^{-y}=C\)
\(\frac{dy}{dx}=e^{x+y}=e^x.e^y\)
\(⇒\frac{dy}{e^y}=e^xdx\)
\(⇒e^{-y} dy=e^x dx\)
Integrating both sides,we get:
\(∫e^{-y} dy=∫e^x dx\)
\(⇒-e^{-y}=e^x+k\)
\(⇒e^x+e^{-y}=-k\)
\(⇒e^x+e^{-y}=c\,\,\, (c=-k)\)
Hence,the correct answer is A.