The solution of the differential equation $\frac{d y}{d x}+1=e^{x+y}$, is

$(x-C) e^{x+y}+1=0$

We have,

$\frac{d y}{d x}+1=e^{x+y}$

Let $x+y=v$. Then, $1+\frac{d y}{d x}=\frac{d v}{d x}$

∴  $\frac{dy}{dx} + 1 = e^{x+y}$

$\Rightarrow \frac{d v}{d x}=e^v$

$\Rightarrow e^{-v} d v=d x$

$\Rightarrow \int e^{-v} d v=\int d x$

$\Rightarrow -e^{-v}=x+C$

$\Rightarrow -e^{-(x+y)}=x+C \Rightarrow(x+C) e^{x+y}+1=0$