Answer
$-e^{-y}$ = $2e^{\sqrt x}+C$
Work Step by Step
$\sqrt x\frac{dy}{dx}$ = $e^{(y+\sqrt x)}$
$e^{-y}dy$ = $\frac{e^{\sqrt x}}{\sqrt x}dx$
$\int(e^{-y})dy$ = $\int\frac{e^{\sqrt x}}{\sqrt x}dx$
$u$ = $\sqrt x$
$du$ = $\frac{1}{2\sqrt x}dx$
$2du$ = $\frac{1}{\sqrt x}dx$
$\int(e^{-y})dy$ = $2\int(e^{u})du$
$-e^{-y}$ = $2e^{u}+C$
$-e^{-y}$ = $2e^{\sqrt x}+C$