It is good recall the following asymptotics. $\cos(x^2) = 1 + \mathcal{O}(x^4)$ $\cos(x) = 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^4)$ $\sin(x^2) = x^2 + \mathcal{O}(x^6)$ $\sin(x) = x + \mathcal{O}(x^3)$
Hence, we get that $\cos(x^2) \cos(x) = \left( 1 + \mathcal{O}(x^4) \right) \left( 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^4) \right) = 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^4)$ $\sin(x^2) \sin(x) = \left(x^2 + \mathcal{O}(x^6) \right) \left( x + \mathcal{O}(x^3) \right) = \mathcal{O}(x^3)$
Hence, we get that $\cos(x^2) \cos(x) + \sin(x^2) \sin(x) = 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^3)$ Hence, $\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x)) = \ln \left(1 - x^2/2 + \mathcal{O}(x^3) \right)$
Also, recall that $\ln(1+t) = t + \mathcal{O}(t^2).$
Hence, $\ln \left(1 - x^2/2 + \mathcal{O}(x^3) \right) = -\dfrac{x^2}{2} + \mathcal{O}(x^3)$ Hence, $\dfrac{\sin(x^2)}{\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x))} = \dfrac{x^2 + \mathcal{O}(x^{6})}{-x^2/2 + \mathcal{O}(x^3)} = \dfrac{-2 + \mathcal{O}(x^4)}{1 + \mathcal{O}(x)}$
Hence, $\lim_{x \to 0} \dfrac{\sin(x^2)}{\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x))} = \lim_{x \to 0} \dfrac{-2 + \mathcal{O}(x^4)}{1 + \mathcal{O}(x)} = \dfrac{\lim_{x \to 0} \left(-2 + \mathcal{O}(x^4) \right)}{\lim_{x \to 0} \left(1 + \mathcal{O}(x) \right)} = -2$
EDIT Below is a slightly different method.
Note that $\cos(x^2) \cos(x) + \sin(x^2) \sin(x) = \cos(x^2 - x)$ We can rewrite $\dfrac{\sin(x^2)}{\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x))}$ as $\dfrac{\sin(x^2)}{\ln(\cos(x^2 - x))} = \dfrac{\sin(x^2)}{x^2} \times \dfrac{x^2}{\ln(\cos(x^2 - x))} = \dfrac{\sin(x^2)}{x^2} \times \dfrac{x^2}{\ln(1 - 2\sin^2((x^2 - x)/2))}$ We used the identity that $\cos(\theta) = 1 - 2 \sin^2 \left( \theta/2\right)$. $\dfrac{\sin(x^2)}{\ln(\cos(x^2 - x))} = \dfrac{\sin(x^2)}{x^2} \times \dfrac{-2\sin^2((x^2 - x)/2)}{\ln(1 - 2\sin^2((x^2 - x)/2))} \times \dfrac{x^2}{-2\sin^2((x^2 - x)/2)}$ Now we have $\lim_{x \to 0} \dfrac{\sin(x^2)}{x^2} = 1$ $\lim_{x \to 0} \dfrac{-2\sin^2((x^2 - x)/2)}{\ln(1 - 2\sin^2((x^2 - x)/2))} = 1$ $\lim_{x \to 0}\dfrac{x^2}{-2\sin^2((x^2 - x)/2)} = -2$ Putting these together again gives us $-2$.