Answer
See graph and explanations.
Work Step by Step
Step 1. Identify the domain of the function: rewrite the function as
$y=\frac{x^3-1+x-1}{-x(x-1)}=\frac{x^2+x+1+1}{-x}=-x-1-\frac{2}{x}, x\ne1$
We can identify the domain as $(-\infty,0)\cup(0,1)\cup (1,\infty)$.
Step 2. Take derivatives to get $y'=-1+\frac{2}{x^2}$ and $y''=-\frac{4}{x^3}$.
Step 3. We can find the possible critical points as $x=0,1,\pm\sqrt 2$ (set $y'=0$). Check the signs of $y'$ across the critical points: $..(-)..(-\sqrt 2)..(+)..(0)..(+)..(1)..(+)..(\sqrt 2)..(-)..$; thus the function decreases on $(-\infty, -\sqrt 2),(\sqrt 2,\infty)$ and increases on $(-\sqrt 2,0),(0,1),(1,\sqrt 2)$. A local minimum can be found at $y(-\sqrt 2)=2\sqrt 2-1$. A local maximum can be found at $y(\sqrt 2)=-2\sqrt 2-1$.
Step 4. Check concavity across $x=0$ with signs of $y''$:$..(+)..(0)..(-)..$; thus the function is concave up on $(-\infty,0)$ and concave up on $(0,\infty)$, but there is no inflection point as the function is not defined at $x=0$..
Step 5. We can identify a vertical asymptote as $x=0$, and a slant asymptote as $y=-x-1$.
Step 6. There are no x- or y-intercepts.
Step 7. Based on the above results, we can graph the function as shown in the figure.
