Solution
= Solution
Near zero, $\cos x=1-x^2/2+O(x^4)$ and $\ln(1+v)=v+O(v^2)$, so $\ln\cos x=-x^2/2+O(x^4)$. Replacing $x$ by $3x$ similarly gives $\ln\cos3x=-9x^2/2+O(x^4)$. Therefore
$$
\boxed{\lim_{x\to0}\frac{\ln\cos x}{\ln\cos3x}=\frac19.}
$$
For all sufficiently small nonzero $x$, both cosines are positive and the denominator logarithm is nonzero, so this quotient and its two-sided real limit are well-defined.