<s>[INST] Let $y= \frac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$ 
Let $y= \dfrac{x^2+3x+1}{x^2+x+1} \forall x\in R$. Find the range of $y$

I have reached: 
$x^2(y-1) + x(y-3) + (y-1) = 0$
I also know that the denominator of $y= f(x)$ is greater than $0$. 
How do I continue from here? I am unable to find a suitable condition to proceed.
Edit: 
vertex of function in the numerator is at $(\frac{-3}{2}, \frac{-5}{4})$
and of the function in the denominator is at: $(\frac{-1}{2},\frac{3}{4})$
but that ain't much helpful, is it? 
 [/INST] You have shown that $y$ is in the range of the function
$$y = \frac{x^2 + 3x + 1}{x + 1}$$
if $x$ is a real-valued root of the quadratic equation
$$x^2(y - 1) + x(y - 3) + (y - 1) = 0$$
For $x$ to be a real-valued root of the quadratic equation, its discriminant must be nonnegative.  The discriminant is
\begin{align*}
\Delta & = b^2 - 4ac\\
       & = (y - 3)^2 - 4(y - 1)(y - 1)\\
       & = y^2 - 6y + 9 - 4(y^2 - 2y + 1)\\
       & = -3y^2 + 2y + 5 
\end{align*}
Hence, we require that 
\begin{align*}
-3y^2 + 2y + 5 & \geq 0\\
3y^2 - 2y - 5 & \leq 0\\
3y^2 + 3y - 5y - 5 & \leq 0\\
3y(y + 1) - 5(y + 1) & \leq 0\\
(3y - 5)(y + 1) & \leq 0
\end{align*}
which holds if $y \in [-1,5/3]$.  Therefore, the range of 
$$y = \frac{x^2 + 3x + 1}{x^2 + x + 1}$$
is 
$$\text{Ran} = \left[-1, \frac{5}{3}\right]$$
</s>