= Solution
Rewrite the <rational function> as $y=1-1/(x^3+1)$. It is undefined at $x=-1$, with the vertical <asymptote> $x=-1$, and tends to the horizontal <asymptote> $y=1$ as $x\to\pm\infty$. At the pole, $y\to+\infty$ from the left and $y\to-\infty$ from the right.
Its <derivative> is
$$
y'=\frac{3x^2}{(x^3+1)^2}\geq0,
$$
vanishing only at $x=0$. Since $y(0)=0$ and $y=x^3+O(x^6)$ near zero, this is a stationary <inflection point>, not an extremum. Thus
$$
\boxed{\text{stationary point }(0,0);\qquad\text{asymptotes }x=-1,\ y=1.}
$$
The graph increases on each side of the vertical <asymptote>. The left branch lies above $y=1$, while the right branch lies below it and crosses the origin horizontally.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2016/paper-1-rational-graph.png]
{title=The rational-function graph, its two asymptotes, and its stationary inflection at the origin}
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