It's easier to sketch if you simplify [tex]f(x)[/tex]. Factorize the numerator and denominator - simple to do with a difference of cubes or squares.
[tex]\dfrac{x^3 - 1}{x^2 - 1} = \dfrac{(x - 1)(x^2 + x + 1)}{(x - 1) (x + 1)}[/tex]
If [tex]x\neq1[/tex], we can cancel the factors of [tex]x-1[/tex]. Note that [tex]f(1)[/tex] itself is undefined. For all intents and purposes, aside from the singularity at [tex]x=1[/tex], the graph of [tex]f(x)[/tex] will look exactly like the graph of
[tex]\dfrac{x^2 + x + 1}{x + 1}[/tex]
Now, rewrite this as
[tex]\dfrac{x(x + 1) + 1}{x + 1}[/tex]
Then when [tex]x\neq-1[/tex] (note that [tex]f(-1)[/tex] is also undefined), we can cancel [tex]x+1[/tex] to reduce this to
[tex]x + \dfrac1{x+1}[/tex]
On its own, the graph of [tex]x[/tex] is a line through the origin. When [tex]x[/tex] is a large number [tex]\frac1{x+1}[/tex] is small. But as [tex]x[/tex] gets closer to -1, the rational term blows up. Effectively, this means the graph of [tex]f(x)[/tex] looks like [tex]\frac1{x+1}[/tex] around [tex]x=-1[/tex], and far enough away it looks like [tex]x[/tex].
See the attached plot for a sketch of these details.
