Note that $\ln(\cos(0))=0$.
So we can write our limit as
$$\lim_{x\to 0^+} \frac{\ln(\cos(x))-\ln(\cos(0))}{(x-0)}.$$
Note that the above expression is almost the usual expression for the derivative of $\ln(\cos(x))$ at $x=0$. (If necessary, go back and look up the definition of the derivative of $f(x)$ at $x=a$). The only difference is the use of a one-sided limit. But what about if the two-sided limit existed?
If we are very lucky and the derivative of $\ln(\cos(x))$ at $0$ exists, the value of that derivative at $x=0$ will be our answer.
So differentiate $\ln(\cos(x))$ in the usual way. Everything works out nicely, the derivative is $0$.
Added: I expect there is no issue in finding the derivative, but here are the details. Using the Chain Rule, we get
$$-(\sin(x))\frac{1}{\cos(x)}.$$
At $x=0$ this is $0$.
So the $0^+$ turns out to be unnecessary, plain old $0$ will do. The manipulations suggested by classmates are not needed, everything follows from the definition of derivative, if we know a couple of differentiation rules.
Comment: This is not really how I would do it, if I needed to know the answer. The "natural" approach is to use the power series expansions of $\cos(x)$ and $\ln(1+u)$. But since you mentioned that you had not yet done L'Hospital's Rule, I assumed that you would not yet have been exposed to power series.
But the power series approach is very much worth knowing. The power series expansion of $\cos x$ is
$$1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!} +\cdots.$$
So very informally, if $x$ is near $0$, $\cos x$ is about $1-x^2/2$.
The power series expansion of $\ln(1+u)$ is
$$u-\frac{u^2}{2}+\frac{u^3}{3} -\frac{u^4}{4}+\cdots.$$
(This expansion is only valid when $-1 \lt u \le 1$.)
So when $x$ is near $0$, $\ln(\cos(x))$ is about $-x^2/2$. Divide by $x$. We get $-x/2$, which approaches $0$ as $x$ approaches $0$.
Best Answer
Hint:
Since $\cos \frac{\pi}{3}=\frac{1}{2}$ we have