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.

Using composition of Taylor series around $x=0$
$$ y=\frac{\sin(x^2)-\sin^2(x)}{x^2\log(\cos (x))}$$
$$\sin(x^2)=x^2-\frac{x^6}{6}+O\left(x^{10}\right)$$
$$\sin^2(x)=x^2-\frac{x^4}{3}+\frac{2 x^6}{45}+O\left(x^8\right)$$
$$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$
$$\log(\cos (x))=-\frac{x^2}{2}-\frac{x^4}{12}+O\left(x^6\right)$$
$$y=\frac{\frac{x^4}{3}-\frac{19 x^6}{90}+O\left(x^8\right) } {-\frac{x^4}{2}-\frac{x^6}{12}+O\left(x^8\right) }=-\frac{2}{3}+\frac{8 x^2}{15}+O\left(x^4\right)$$ shows the limit and how it is approached.
