= Solution
The <domain of a function> is $\mathbb R\setminus\{0\}$, because the denominator vanishes at zero. Near zero, $e^x-1\sim x$, so $x=0$ is a vertical <asymptote>, approached with value $-\infty$ from the left and $+\infty$ from the right. The horizontal <asymptotes> are $y=-1$ as $x\to-\infty$ and $y=0$ as $x\to+\infty$.
Moreover,
$$
f'(x)=-\frac{e^x}{(e^x-1)^2}<0.
$$
Both branches define a strictly decreasing <monotone function>: the left branch lies below $-1$, and the right branch is positive. There are no zeros or vertical-axis intercept.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2012/paper-1-exponential-reciprocal.png]
{title=The decreasing reciprocal-exponential branches with their three asymptotes}
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