2. (D).

By Vieta's formulas, we get $a+c=-\frac{b}{a}, a c=\frac{c}{a}$, then $a^{2} c=$ c. Since $a c \neq 0$, we have $a= \pm 1$. The equation $9 c x^{2}+3 b x+a=0(a c$ $\neq 0$ ) has no zero roots, so,
$$c+\frac{b}{3 x}+\frac{a}{9 x^{2}}=0,$$

which means $a \cdot\left(\frac{1}{3 x}\right)^{2}+b \cdot\left(\frac{1}{3 x}\right)+c=0$.
Therefore, when $a=1$, we have $\frac{1}{3 x}=1, x=\frac{1}{3}$;
when $a=-1$, $\frac{1}{3 x}=-1, x=-\frac{1}{3}$.