Calculating $\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^2\ln(\cos x)}$ without L'Hospital's rule this is my first post here so excuse the lack of knowledge about how things usually go.
My question revolves around calculating the limit as $x$ approaches $0$ of the following function:
$$\lim_{x \to 0} \frac{\sin(x^2)-\sin^2(x)}{x^2\ln(\cos x)}$$
The question came up in a test about a month ago and while I couldn't solve it in the test I've been working on it since then but I can't seem to get it. I know the limit is supposed to be $\frac{-2}{3}$ from some online calculators which abused l'hopital rule over and over again. I've tried playing around with it in so many ways but I always seem to get 0 over 0 or the so called indeterminate form. I've even tried calculating it by substituting in the Taylor series for the functions given but no luck. If anyone could show me a method of calculating this without using l'hopital rule or better yet, give me a hint as to how I should proceed I would be grateful.

HINT
Use the following Maclaurin series:
$$\begin{align}
\sin x&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots\\
\ln(1-x)&=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots\end{align}$$
along with the trig identity $\cos x\equiv1-2\sin^2\frac{x}{2}$.
