I'm trying to find out if this limit exists

You can also just use L'Hôpital's rule directly and the 2nd fundamental theorem of calculus: since $$ \frac{d}{dx} \int_0^{x^2} \sin \sqrt{t} \ dt = \sin(x)\cdot2x, $$ the indeterminate limit is equal to $$ \lim_{x \to 0} \frac{\sin(x)\cdot2x}{3x^2} = \frac{2}{3} \lim_{x \to 0} \frac{\sin(x)}{x}. $$ The latter limit is known to be $1$ (from the standard proof that the derivative of the sine function is the cosine function).

$$\begin{align*} \lim_{x \to 0^+} \frac{1}{x^3} \int_0^{x^2} \sin \sqrt{t} \ \mathrm dt &= \lim_{x \to 0^+} \frac{1}{x^3} \int_0^{x^2} \sqrt{t} \left( 1 + O(t) \right) \ \mathrm dt \\ &= \lim_{x \to 0^+} \frac{1}{x^3} \left( \frac{2}{3} t^{3/2} + O(t^{5/2}) \right)_0^{x^2} \\ &= \lim_{x \to 0^+} \frac{1}{x^3} \left( \frac{2}{3} x^3 + O(x^5) \right) \\ &= \lim_{x \to 0^+} \left(\dfrac23+O(x^2)\right) \\ &= \frac{2}{3} \end{align*}$$

Letting $t=u^2$, we get $$ 2\int_0^{x} u \sin u\, du\ . $$ Integrating by parts, $$ \int_0^{x} u \sin u \, du = -x \cos x + \int_0^x \cos u\, du = \sin x -x\cos x. $$ Hence we get, by applying de l'Hôpital's rule, $$ \lim_{x\to0}\frac{2}{x^3}\left( \sin x - x\cos x \right)= \lim_{x\to0}\frac{2x \sin x}{3 x^2}=\frac{2}{3}. $$