Firstly, we can use chain rule, which is:
$\frac{d}{dx}f(g(x))=f'(g(x))g'(x)$
where $f(x)=e^x$ and $g(x)=-x$ (so that $f(g(x))=e^{-x}$). We know that $\frac{d}{dx}e^x=e^x$ and $\frac{d}{dx}-x=-1$ so we get the derivative:
$\frac{d}{dx}e^{-x}=e^{-x}*-1=-e^{-x}$
To know why this occurs, we use first principles:
$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$
so thus
$\frac{d}{dx}e^{-x}=\lim_{h\rightarrow 0}\frac{e^{-(x+h)}-e^{-x}}{h} = \lim_{h\rightarrow 0}e^{-x}(\frac{e^{-h}-1}{h})$
now, from the (a?) definition of $e$:
$\lim_{h\rightarrow 0}\frac{e^{-h}-1}{h}=-1$
and thus
$\frac{d}{dx}e^{-x}=-e^{-x}$