Solution to Sextic Polynomial with Two Real Roots

Actually, your sextic has a solvable Galois group so can be exactly expressed in terms of radicals. The solution turns out to be quite simple. Given

$$3a x ^6 + 6x^5 - 9a x^4 - 4x^3 + 9a x^2 + 6x - 3a = 0\tag1$$

Then the two real roots are,

$$x = u\pm\sqrt{u^2+1}\tag2$$

where,

$$u=\frac{1}{3a}\Big(z_1^{1/3}+\frac{1}{z_1^{1/3}}-1\Big)\tag3$$

and $z_1$ is any non-zero root of,

$$z^2+(9a^2+2)z+1=0\tag4$$

Your sextic's positive real root $x$ is the positive case $\pm$ of $(2)$.

Note: Be careful when testing this numerically, especially using $(3)$. What I find supremely annoying about my version of Mathematica is it gives a complex value to $z^{1/3}$ when $z$ is a negative real number.

P.S.: However, an easy way to verify this is to use resultants and the Mathematica syntax,

Factor[Resultant[x^2 - 2u x - 1, a u^3 + u^2 + 1/3, u]]

which then yields $(1)$, showing that your sextic in just a quadratic $(2)$ in $x$, with coefficients that is cubic $(3)$ in $u$.