Find $\lim_{x\to0}\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{x^2}$. Is my approach correct?

In a slightly different way, using the Taylor expansion, as $x \to 0$, $$ \sin x=x-\frac{x^3}6+O(x^5) $$ gives $$ 1-\frac{\sin x}x=\frac{x^2}6+O(x^4) $$ then $$ \sin \left( 1-\frac{\sin x}x\right)=\frac{x^2}6+O(x^4) $$ and

$$ \frac{\sin \left( 1-\frac{\sin x}x\right)}{x^2}=\frac16+O(x^2) $$

from which one may conclude easily.

By L' Hospital anyway:

$$\frac{\sin\left(1-\frac{\sin(x)}{x}\right)}{x^2}$$ yields

$$\cos\left(1-\frac{\sin(x)}x\right)\frac{\sin(x)-x\cos(x)}{2x^3}.$$

The first factor has limit $1$ and can be ignored.

Then with L'Hospital again:

$$\frac{x\sin(x)}{6x^2},$$

which clearly tends to $\dfrac16$.