Answer
(a) $x=0$ is a vertical asymptote.
$y = 0$ is a horizontal asymptote.
$y = -1$ is a horizontal asymptote.
(b) $f$ is increasing on the intervals $(-\infty,0)\cup(0,\infty)$
(c) There is no local maximum or local minimum.
(d) The graph is concave down on this interval: $(0, \infty)$
The graph is concave up on this interval: $(-\infty,0)$
There are no points of inflection.
(e) We can see a sketch of the graph below.
Work Step by Step
(a) $f(x) = \frac{e^x}{1-e^x}$
$f(x)$ is undefined when $1-e^x = 0$
$x=0$ is a vertical asymptote.
$\lim\limits_{x \to -\infty}f(x) = 0$
$y = 0$ is a horizontal asymptote.
$\lim\limits_{x \to \infty}f(x) = -1$
$y = -1$ is a horizontal asymptote.
(b) We can find the points where $f'(x) = 0$:
$f'(x) = \frac{(e^x)(1-e^x)-(e^x)(-e^x)}{(1-e^x)^2}$
$f'(x) = \frac{e^x-e^{2x}+e^{2x}}{(1-e^x)^2}$
$f'(x) = \frac{e^x}{(1-e^x)^2}$
There are no values of $x$ such that $f'(x) = 0$
Note that $f'(x)$ is undefined when $x=0$
$f$ is not decreasing on any interval.
When $x \lt 0$ or $x \gt 0$ then $f'(x) \gt 0$
$f$ is increasing on the intervals $(-\infty,0)\cup(0,\infty)$
(c) There are no values of $x$ such that $f'(x) = 0$
There is no local maximum or local minimum.
(d) We can find the points where $f''(x) = 0$:
$f''(x) = \frac{(e^x)(1-e^x)^2-(e^x)(2)(1-e^x)(-e^x)}{(1-e^x)^4}$
$f''(x) = \frac{(e^x)(1-2e^x+e^{2x})+(2e^{2x})(1-e^x)}{(1-e^x)^4}$
$f''(x) = \frac{e^x-2e^{2x}+e^{3x}+2e^{2x}-2e^{3x}}{(1-e^x)^4}$
$f''(x) = \frac{e^x-e^{3x}}{(1-e^x)^4}$
$f''(x) = \frac{(1-e^x)(e^{2x}+e^x)}{(1-e^x)^4}$
$f''(x) = \frac{e^{2x}+e^x}{(1-e^x)^3}$
There are no values of $x$ such that $f''(x) = 0$
Note that $f''(x)$ is undefined when $x=0$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on this interval: $(0, \infty)$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on this interval: $(-\infty,0)$
There are no points of inflection.
(e) We can see a sketch of the graph below.