$$f(x)=(x^{2}+x+1) \cdot e^{-x}$$
$$f'(x)=e^{-x}(x-x^{2})$$
$$f''(x)=e^{-x}(x^{2}-3x+1)$$
(Both derivatives are correct, checked this with an online calculator.)
$$f'(x)=0$$
$$0=e^{-x}(x-x^{2})$$
$$0=x-x^{2}$$
$$0=x(1-x)$$
$$x_{1}=0$$
$$x_{2}=1$$
Check if these are maximum / minimum:
$$f''(0)=1>0$$
$\Rightarrow$ minimum at $P(0|1)$
$$f''(1) = -\frac{1}{e}<0$$
$\Rightarrow$ maximum at $Q(1|\frac{3}{e})$
Now check if local or global extremums:
$$\lim_{x\rightarrow\infty}(x^{2}+x+1) \cdot e^{-x}= 0$$
$$\lim_{x\rightarrow -\infty}(x^{2}+x+1) \cdot e^{-x}= \infty$$
Thus $P$ is a global minimum and $Q$ is a local maximum.
I'd like to know if everything is alright? The bolt printed means this is the part I care most about, the main reason I ask this question. Because I'm not sure if this is right, about the rest, it seems correct to me but who knows :o
$P$ is not a global minimum because, as you stated, the limit of the function as $x\to \infty$ is 0. There is a value of $x$ where $f(x)<1 = P(0).$
And to check the rest of your work: Plot of $f(x)$