Answer
$y = -ln(x^2/2 + C)$
Work Step by Step
$y' +xe^y = 0$ can be rewritten as :
$dy/dx + xe^y = 0$
Now let's separate the y on one side and the x on the other side:
$dy/dx = -xe^y$
$e^{-y}dy = -xdx$
**It is to be noted that $1/e^y = e^{-y}$**
Now let's integrate both sides:
$\int e^{-y}dy =\int -xdx$
$-e^{-y} = -x^2/2 + C$
$e^{-y} = x^2/2 + C_2$
To isolate y, we need to add ln() to both sides:
$ln(e^{-y}) = ln(x^2/2 + C_2)$
$-y = ln(x^2/2 + C_2)$
$y = -ln(x^2/2 + C_2)$