Since you have sorted out the source of misunderstanding, here I offer an alternative solution, which quite surprisingly* allows you to find the exact roots of the polynomial while avoiding calculus. The method is (much) longer, but I find it interesting and I hope you will too.
(*I say surprisingly because mostly when solving quartic equations we don't have a hope of finding the exact roots without using the quartic formula.)
Alternative Solution
We begin with a key observation: when a polynomial's coefficients are mirror images of each other about the centre of the polynomial, then the roots of the polynomial come as pairs of reciprocals of each other. For example, the roots of the polynomial $2x^2+5x+2$ are $-2$ and $-0.5$ which are reciprocals of each other.
If you would like me to, I can prove this for you, but it may be a good exercise to try it for yourself.
Using this fact, we may now observe that your polynomial satisfies this property: the coefficients are unchanged if we 'reflect' them about the middle term.
Therefore, since the polynomial is a quartic it has $4$ roots and its roots must be of the form
$$-\alpha,-\frac{1}{\alpha},-\beta, -\frac{1}{\beta}.$$
Using the Factor theorem, we can now see that your polynomial can be factorised in the following way:
$$\begin{align}3x^4+6x^3+x^2+6x+3&\equiv 3(x+\alpha)\left(x+\frac{1}{\alpha}\right)(x+\beta)\left(x+\frac{1}{\beta}\right)\\
&\equiv 3\left(x^2+x\left(\alpha+\frac{1}{\alpha}\right)+1\right)\left(x^2+x\left(\beta+\frac{1}{\beta}\right)+1\right)\\
&\equiv 3x^4+3x^3\left(\alpha+\frac{1}{\alpha}+\beta+\frac{1}{\beta}\right)+3x^2\left(2+\left(\alpha+\frac{1}{\alpha}\right)\left(\beta+\frac{1}{\beta}\right)\right)\\
&\quad+3x\left(\alpha+\frac{1}{\alpha}+\beta+\frac{1}{\beta}\right)+3
\end{align}$$
Divide both sides by $3$ and for simplicity's sake let
$$a=\alpha+\frac{1}{\alpha},\quad b=\beta+\frac{1}{\beta}$$
and we find that
$$x^4+2x^3+\frac{1}{3}x^2+2x+1\equiv x^4+x^3(a+b)+x^2(2+ab)+x(a+b)+1.$$
What we really have here is a pair of simultaneous equations:
$$\begin{align}
a+b&=2\iff a=2-b\\
2+ab&=\frac{1}{3}
\end{align}$$
Replacing the $a$ in the $2$nd equation by the $a$ in the $1$st equation, we are left with a quadratic in $b$:
$$\begin{align}
2+b(2-b)=\frac{1}{3}&\implies b^2-2b-\frac{5}{3}=0\\
&\implies b=\frac{3\pm 2\sqrt6}{3}
\end{align}$$
The nice thing about this solution is that we then find that
$$a=\frac{3\mp 2\sqrt6}{3}.$$
Therefore, without loss of generality we can let
$$a=\frac{3+2\sqrt6}{3},\quad b=\frac{3- 2\sqrt6}{3}.$$
We now have $2$ quadratic equations that allow us to find the exact roots of the quartic polynomial!!
$$\begin{align}&\frac{3+2\sqrt6}{2}=\alpha+\frac{1}{\alpha}\implies \alpha^2-\alpha\left(\frac{3+ 2\sqrt6}{2}\right)+1=0\\
&\frac{3-2\sqrt6}{2}=\beta+\frac{1}{\beta}\implies \beta^2-\beta\left(\frac{3-2\sqrt6}{2}\right)+1=0
\end{align}$$
Now all you need to find the values of $-\alpha,-\frac{1}{\alpha},-\beta, -\frac{1}{\beta}$ is to solve these two quadratic equations, and indeed only one of them has real roots (which you can find exactly using the quadratic formula).
I hope that was helpful, or you at least found it interesting. Sorry for the length, I've just tried to make it is clear as possible. If you have any questions please don't hesitate to ask :)
