Differentiating $e^x + e^y = e^{x + y}$ Differentiating with respect to x, $e^x + e^y = e^{x + y}$, could anyone give me a hint? I do not know even how to start, taking the ln of both sides does not solve the problem.   

You didn't provide enough information, so I am going to assume that you are taking the derivative with respect to $x$, that $y = y(x)$, and that are familiar with the chain rule. 
Let $f(x) = e^x$, so $f'(x)= e^x $
$\frac{d}{dx}f(y(x)) = f'(y(x))y'(x) = e^y \frac{dy}{dx} $
$\frac{d}{dx}f(x+y(x)) = f'(x+y(x))(1 + y'(x)) = e^{x+y}(1+ \frac{dy}{dx}) $
So, after differentiating, the equation becomes
$$e^x+e^y \frac{dy}{dx} = e^{x+y}(1+ \frac{dy}{dx}) $$
