Sketch the graph of the function $f(x) =\frac{(x+1)^3} {x^2 - x+1}$ I tried to follow the rules of sketching a graph, I found the intersection with the ox and oy in (0,1)and (-1,0) and I think that the oblique asymptote is y=x+4. What about the vertical ones? The roots of the denominator are complex ones. Can we say these are asymptotes?.. The first derivative is pretty hard to find roots of. How can I sketch its graph, please?

You have $$f(x) =\frac{(x+1)^3} {x^2 - x+1} = x + 4 + \frac{6x - 3}{x^2-x+1}.$$
As $\lim\limits_{x \to \pm \infty} \frac{6x - 3}{x^2-x+1} = 0$, $y=x+4$ is indeed an asymptote of $f$. Also the sign of $\frac{6x - 3}{x^2-x+1}$ is the one of $6x-3$ as $x^2-x+1$ is always positive. So $f$ crosses its asymptote at $x = 1/2$, is above it for $x > 1/2$ and below it for $x < 1/2$.
You can also find where the maximal vertical distance between $f$ and its asymptote is by vanishing the derivative of $g(x)=\frac{6x - 3}{x^2-x+1}$. You'll find two symmetric values about $x=1/2$.
Some elements to start a decent plot...
