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 $0$.
As mentioned your first solution is incorrect. the reason is $$\displaystyle lim_{x\to0}\frac{2xcos(2x)}{sin(2x)}\neq 0$$ you can activate agin l'hospital:
$$lim_{x\to0}\frac{2xcos(2x)}{sin(2x)}=lim_{x\to0}\frac{2cos(2x)-2xsin(2x)}{2cos(2x)}=lim_{x\to0} 1-2xtg(2x)=1+0=1$$, so now after we conclude this limit, in the first solution, i'd write after $$\lim\limits_{x\to 0}\left(\frac{1+x\sin(2x)-\cos(x)}{\sin^2(x)}\right)=\lim\limits_{x\to 0}\frac{\sin(2x)}{\sin(2x)}+\frac{2x\cos(2x)}{\sin(2x)}+\frac{\sin(x)}{\sin(2x)}$$ that the limit equals to $$ = 1+\frac 1 2+1=\frac 5 2$$ and that's the correct answer.
Best Answer
The following steps use De l'Hopital rule.
$$\lim _{x\to 0}\left(\frac{e^{-\frac{x^2}{2}}-\cos \left(x\right)}{x^3\sin \left(x\right)}\right)=\lim _{x\to \:0}\left(\frac{-e^{-\frac{x^2}{2}}x+\sin \left(x\right)}{3x^2\sin \left(x\right)+\cos \left(x\right)x^3}\right)=\lim _{x\to \:0}\left(\frac{e^{-\frac{x^2}{2}}x^2-e^{-\frac{x^2}{2}}+\cos \left(x\right)}{-x^3\sin \left(x\right)+6x^2\cos \left(x\right)+6x\sin \left(x\right)}\right)$$ $$=\lim _{x\to 0}\left(\frac{-e^{-\frac{x^2}{2}}x^3+3e^{-\frac{x^2}{2}}x-\sin \left(x\right)}{-x^3\cos \left(x\right)-9x^2\sin \left(x\right)+18x\cos \left(x\right)+6\sin \left(x\right)}\right)=\lim _{x\to 0}\left(\frac{e^{-\frac{x^2}{2}}x^4-6e^{-\frac{x^2}{2}}x^2+3e^{-\frac{x^2}{2}}-\cos \left(x\right)}{x^3\sin \left(x\right)-12x^2\cos \left(x\right)-36x\sin \left(x\right)+24\cos \left(x\right)}\right)$$ $$=\frac{e^{-\frac{0^2}{2}}\cdot 0^4-6e^{-\frac{0^2}{2}}\cdot 0^2+3e^{-\frac{0^2}{2}}-\cos \left(0\right)}{0^3\sin \left(0\right)-12\cdot 0^2\cos \left(0\right)-36\cdot 0\cdot \sin \left(0\right)+24\cos \left(0\right)}=\frac 1{12}. \square$$