If the graph of the continuous function y = f(x) passes through (a, 0), then $\lim\limits_{x \rightarrow a} \frac{\ln \left(1+6 f^2(x)-3 f(x)\right.}{3 f(x)}$ is equal to:

-1

Since $f(a)=0 \Rightarrow \lim\limits_{x \rightarrow a}\left(6 f^2(x)-3 f(x)\right)=0$

$\lim\limits_{x \rightarrow a} \frac{\ln \left(1+6 f^2(x)-3 f(x)\right)}{3 f(x)}=\lim\limits_{x \rightarrow a} \frac{\ln \left(1+6 f^2(x)-3 f(x)\right)}{\left(6 f^2(x)-3 f(x)\right)} . \frac{\left(6 f^2(x)-3 f(x)\right)}{3 f(x)}$

$=\lim\limits_{x \rightarrow a} \frac{6 f(x)-3}{3}=-1$

Hence (3) is the correct answer.