Did I do everything correctly?
Find all local extremums of the following function and decide if these are global extremums (i.e. maxima or rather minima of the function on its entire domain) or not:
$f:\mathbb{R}\ni x\mapsto\frac{x}{x^{2}+x+1} \in \mathbb{R}$
Check behavior of function $f$ for $x\rightarrow +- \infty$ (because maybe there will be global extremums but they could be the limit..):
$$\lim_{x\rightarrow+\infty}f(x)=0$$
$$\lim_{x\rightarrow-\infty}f(x)=-0$$
$$f'(x) = -\frac{x^{2}+1}{(x^{2}+x+1)^{2}}$$
$$f'(x)=0$$
$\Rightarrow$
$$0=-\frac{x^{2}+1}{(x^{2}+x+1)^{2}}$$
$$0=\frac{x^{2}+1}{(x^{2}+x+1)^{2}}$$
$$0=x^{2}+1$$
$$x^{2}=-1$$
There is no result which means that the function $f$ hasn't got any extremums.
Your limit as of $f$ as $x \to \pm \infty$ is all good, however, your derivative has a little snag in it. Using the quotient rule we have $$f'(x) = \frac{(x^2 + x+1) - x(2x +1)}{(x^2 + x+1)^2} = \frac{1 - x^2}{(x^2+x+1)^2}$$ and so $f'$ vanishes precisely at $x = \pm 1$, giving $f(-1) = 1$ a global maximum and $f(1) = \frac{1}{3}$ a global minimum.
Note: Since your $f$ is defined everywhere and tends to $0$ as $x \to \pm \infty$ it must have at least one global extrema. Think of the shape of the graph or the IVT for intuition and rigor respectively.