Solution
= Solution
Apply <implicit differentiation>, remembering that $y$ depends on $x$:
$$
1+y'+e^x+e^y y'=0.
$$
Since $1+e^y>0$, division is always legitimate. Therefore
$$
\boxed{\frac{dy}{dx}=-\frac{1+e^x}{1+e^y}.}
$$
= Solution
Apply <implicit differentiation>, remembering that $y$ depends on $x$:
$$
1+y'+e^x+e^y y'=0.
$$
Since $1+e^y>0$, division is always legitimate. Therefore
$$
\boxed{\frac{dy}{dx}=-\frac{1+e^x}{1+e^y}.}
$$