Given $\large\frac{dy}{dx}$$+1=e^{x+y}$-----(1)
Put $x+y=z$ on differentiating w.r.t x we get,
$1+\large\frac{dy}{dx}=\frac{dz}{dx}$$=>\large\frac{dy}{dx}=\frac{dz}{dx}$$-1$
Now substituting in equ(1) we get
$\large\frac{dz}{dx}$$-1+1=e^z$
=>$ \large\frac{dz}{dx}$$=e^z$
Now seperating the variables we get
$\Large\frac{dz}{e^z}$$=dx$
=>$e^{-z}dz=dx$
On integrating we get: $ \int e^{-z}dz=\int dx$
$-e^{-z}=x+c$
Substituting for z we get
$-e^{-(x+y)}=x+c$
Dividing throughout by $e^{-(x+y)}$ we get
$-1=e^{(x+y)}(x+c)$
Therefore $(x+c)e^{(\large x+y)}$$+1=0$