Squeeze Theorem.
First, there is some factoring and cancellation: $ \begin{align} \frac{x^2-1+\cos^2x}{x^4+x^3\sin x} &=\frac{x^2-(1-\cos^2{x})}{x^3(x+\sin x)}\\ &=\frac{x^2-\sin^2(x)}{x^3(x+\sin x)}\\ &=\frac{(x-\sin x )(x+\sin x)}{x^3(x+\sin x)}\\ &=\frac{x-\sin x}{x^3}\\ \end{align} $
The function $f$ with $f(x)=\frac{1}{6}x^3$ and $g$ with $g(x)=x-\sin x$ satisfy the inequality $g(x)\leq f(x)$ for all $x\in(0,\epsilon)$. This is because at $x=0$, the two functions have the same value, derivative, second derivative, third derivative, and fourth derivative, but $g^{(5)}(0).
And if $h(x)=\frac{1}{6}x^3 - x^5$, then $h(x)\leq g(x)$ in $(0,\epsilon)$ for basically the same reason, but now $h^{(5)}(0).
So for all $x\in (0,\epsilon)$, $h(x)\leq g(x)\leq f(x)$ $\implies\frac{h(x)}{x^3}\leq \frac{g(x)}{x^3}\leq \frac{f(x)}{x^3}$ $\implies\frac{1}{6}-x^2\leq \frac{x-\sin x}{x^3}\leq \frac{1}{6}$
Applying the Squeeze Theorem as $x\to0^+$ gives that $\lim_{x\to0^+}\frac{x-\sin x}{x^3}=\frac{1}{6}$ You can either alter this argument to work on both sides at once, or make a separate similar argument for the other side.