Here is a different approach that uses the integral definition of the arcsine function. We will deduce the limit of interest, $\lim_{h\to 0}\frac{\sin(h)}{h}=1$, without appeal to geometry or differential calculus. (Note that $\cos(h)-1=-2\sin^2(h/2)$)
Instead, we only rely on elementary analysis of continuous functions and their inverses along with simple properties of the Riemann integral. To that end, we now proceed.
We define the sine function, $\sin(x)$, as the inverse function of the function $f(x)$ given by
$$\bbox[5px,border:2px solid #C0A000]{f(x)=\int_0^x \frac{1}{\sqrt{1-t^2}}\,dt }\tag 1$$
for $|x|< 1$.
NOTE: It can be shown that the sine function defined as the inverse of $f(x)$ given in $(1)$ has all of the familiar properties that characterize the circular function $\sin(x)$.
It is straightforward to show that since $\frac{1}{\sqrt{1-t^2}}$ is positive and continuous for $t\in (-1,1)$, $f(x)$ is continuous and strictly increasing for $x\in (-1,1)$ with $\displaystyle\lim_{x\to 0}f(x)=f(0)=0$.
Therefore, since $f$ is continuous and strictly increasing, its inverse function, $\sin(x)$, exists and is also continuous and strictly increasing with $\displaystyle \lim_{x\to 0}\sin(x)=\sin(0)=0$.
From $(1)$, we have the bounds (SEE HERE)
$$\bbox[5px,border:2px solid #C0A000]{1 \le \frac{f(x)}x\le \frac{1}{\sqrt{1-x^2}}} \tag 2$$
for $x\in (-1,1)$, whence applying the squeeze theorem to $(2)$ yields
$$\lim_{x\to 0}\frac{f(x)}{x}=1 \tag 3$$
Finally, let $y=f(x)$ so that $x=\sin(y)$. As $x\to 0$, $y\to 0$ and we can write $(3)$ as
$$\lim_{y\to 0}\frac{y}{\sin(y)}=1$$
from which we have
$$\bbox[5px,border:2px solid #C0A000]{\lim_{y\to 0}\frac{\sin(y)}{y}=1}$$
as was to be shown!
NOTE:
We can deduce the following set of useful inequalities from $(2)$. We let $x=\sin(\theta)$ and restrict $x$ so that $x\in [0,1)$. In addition, we define new functions, $\cos(\theta)=\sqrt{1-\sin^2(\theta)}$ and $\tan(\theta)=\sin(\theta)/\cos(\theta)$.
Then, we have from $(2)$
$$\bbox[5px,border:2px solid #C0A000]{y\cos(y)\le \sin(y)\le y\le \tan(y)} $$
which are the familiar inequalities often introduced in an introductory geometry or trigonometry course.
$$\left( \mathrm M (t + \mathrm d t) \right)^{-1} = \left( \mathrm M (t) + \dot{\mathrm M} (t) \,\mathrm d t \right)^{-1} = \cdots = \mathrm M^{-1} (t) \color{blue}{- \mathrm M^{-1} (t) \, \dot{\mathrm M} (t) \, \mathrm M^{-1} (t)} \,\mathrm d t$$
Best Answer
You didn't try hard enough. Just for curiosity's sake, here you go. Notice this follows the pattern of the usual proof of the derivative of $\sin$, using the addition formula for $\sin$. First, \begin{align*} \lim_{h\to 0}&\frac{\sin((x+h)^2) - \sin(x^2)}h \\ &= \sin(x^2)\lim_{h\to 0}\frac{\cos(2xh+h^2)-1}h + \cos(x^2)\lim_{h\to 0}\frac{\sin(2hx+h^2)}h. \end{align*} Now, recall from your text that $$\lim_{h\to 0}\frac{\sin h}h = 1,$$ so we have $$\lim_{h\to 0}\frac{\sin(2xh+h^2)}h = \lim_{h\to 0}\frac{\sin(2xh+h^2)}{2xh+h^2}\lim_{h\to 0}\frac{2xh+h^2}h = 1\cdot 2x.$$ Similarly, you should recall from your text that $$\lim_{h\to 0}\frac{\cos(h)-1}h = 0,$$ and so the same approach will show that the first term goes to $0$. This gives $$\lim_{h\to 0}\frac{\sin((x+h)^2) - \sin(x^2)}h = 2x\cos(x^2),$$ as you desired.
COMMENT: This is essentially how the chain rule is proved, by the way, if you don't worry about every last technicality.