Answer
A. The domain is $(-\infty,0)\cup (0,\infty)$
B. There is no y-intercept.
The x-intercept is $1$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} (\frac{x-1}{x^2}) = 0$
$\lim\limits_{x \to \infty} (\frac{x-1}{x^2}) = 0$
$y = 0$ is a horizontal asymptote.
$x = 0$ is a vertical asymptote.
E. The function is decreasing on the intervals $(-\infty, 0)\cup (2, \infty)$
The function is increasing on the interval $(0, 2)$
F. The local maximum is $(2, \frac{1}{4})$
G. The graph is concave down on the intervals $(-\infty,0)\cup (0,3)$
The graph is concave up on the interval $(3, \infty)$
The point of inflection is $(3, \frac{2}{9})$
H. We can see a sketch of the curve below.

Work Step by Step
$y = \frac{x-1}{x^2}$
A. The function is defined for all real numbers except $x=0$
The domain is $(-\infty,0)\cup (0,\infty)$
B. There is no y-intercept since $x \neq 0$
When $y = 0$:
$\frac{x-1}{x^2} = 0$
$x-1 = 0$
$x = 1$
The x-intercept is $1$
C. The function is not an even function or an odd function.
D. $\lim\limits_{x \to -\infty} (\frac{x-1}{x^2}) = 0$
$\lim\limits_{x \to \infty} (\frac{x-1}{x^2}) = 0$
$y = 0$ is a horizontal asymptote.
E. We can find the values of $x$ such that $y' = 0$:
$y' = \frac{(1)(x^2)-(x-1)(2x)}{x^4}$
$y' = \frac{x^2-2x^2+2x}{x^4}$
$y' = \frac{2x-x^2}{x^4}$
$y' = \frac{2-x}{x^3} = 0$
$2-x = 0$
$x = 2$
When $x \lt 0$ or $x \gt 2$ then $y' \lt 0$
The function is decreasing on the intervals $(-\infty, 0)\cup (2, \infty)$
When $0 \lt x \lt 2$, then $y' \gt 0$
The function is increasing on the interval $(0, 2)$
F. When $x = 2$ then $y = \frac{(2)-1}{(2)^2} = \frac{1}{4}$
The local maximum is $(2, \frac{1}{4})$
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = \frac{(-1)(x^3)-(2-x)(3x^2)}{x^6}$
$y'' = \frac{-x^3-6x^2+3x^3}{x^6}$
$y'' = \frac{2x^3-6x^2}{x^6}$
$y'' = \frac{2x-6}{x^4} = 0$
$2x-6 = 0$
$x = 3$
When $x \lt 0~~$ or $0 \lt x \lt 3$, then $y'' \lt 0$
The graph is concave down on the intervals $(-\infty,0)\cup (0,3)$
When $x \gt 3$, then $y'' \gt 0$
The graph is concave up on the interval $(3, \infty)$
When $x = 3,$ then $y = \frac{(3)-1}{(3)^2} = \frac{2}{9}$
The point of inflection is $(3, \frac{2}{9})$
H. We can see a sketch of the curve below.
