The value of $x$ for which function attains max value 
At what value of $x,\ x\in \mathbb{Z}$ will the function $\dfrac{x^2+3x+1}{x^2-3x+1}$ attain its maximum value .

$\color{green}{a.)\ 3 }\\
b.)\ 4 \\
c.) -3 \\
d.)\ \text{none of these} \\ $
$\dfrac{x^2+3x+1}{x^2-3x+1}\\
=1+\dfrac{6x}{x^2-3x+1}\\
=1+\dfrac{6}{x-3+\frac{1}{x}}\\
$
here i thought to use $\text{AM-GM}$ inequality for the part $x+\dfrac{1}{x}$
and concluded that $x+\dfrac{1}{x}=2 \implies x=1$ 
But after inspecting some values i  came up with
$$\begin{array}{|c|c|} \hline
x & k  \\ \hline
-4 & \dfrac{5}{29}  \\ \hline
-3 & \dfrac{1}{19}  \\ \hline
-2 & -\dfrac{1}{11}  \\ \hline
-1 & -\dfrac{1}{5}  \\ \hline
0 & 1 \\ \hline
1 & -5 \\ \hline
2 & -11 \\ \hline
\color{green}{3 }& \color{red}{19} \\ \hline
4 &\dfrac{29}{5}\\ \hline
\end{array}$$
I look for a short and simple way .
also i don't want to use calculus.
I have studied maths up to $12$th grade.
 A: Your last two formulas for $f(x)$ in your first edit are incorrect: they should be
$$f(x)=1+\frac{6x}{x^2-3x+1}=1+\frac{6}{x-3+\frac{1}{x}}$$
(I see you corrected that formula in your question by a later edit.)
You maximize that by making the denominator positive but as small as possible. The $-3$ in that denominator means that you make $x=3$. Any smaller, and the denominator becomes negative. Any bigger, and the denominator becomes bigger and reduces the value of the function.
Nothing fancy was needed here. The AM-GM inequality shows that $x+\frac 1x$ has a minimum of $2$, but that makes the denominator negative and is thus not relevant here. You want to make $x+\frac 1x$ just above $3$, and that is done at $x=3$.

Here is another way to look at it. If $x$ can be any real number, your function $f(x)$ has no maximum. It tends to infinity as the denominator tends to zero. We can solve the quadratic equation of the denominator equaling zero and get
$$x=\frac{3\pm\sqrt 5}{2}\approx 0.38,\ 2.62$$
We get a positive denominator for $x<0.38$ and $x>2.62$ (approximations used here).
We still want a small positive denominator, so if we limit $x$ to integers we now look at $x=0$ (below 0.38) and $x=3$ (above 2.62). We see that $x=3$ gives us the larger value, so that gives us the maximum for all integers.
