Derivative of $-e^y = 0$?

Question

I stumbled upon this on wolfram alpha and still wonder why $-e^x$ equals $0$ (third step).

Answer 1

It is assuming that $x$ and $y$ are independent variables and as you're differentiating with respect to $y$, that term vanishes since it doesn't contain any $y$'s

Answer 2

Variable $x$ is treated like a constant (becouse you are differentiating with respect to $y$). So $$(-e^x)'=(-e^C)'=(C)'=0.$$