Suppose $f$ is even and infinitely differentiable, and let $g(x) = f(\sqrt{x})$. I claim that for each positive integer $k$, $g^{(k)}$ is defined on $[0, \infty)$, and moreover that $g^{(k)}(x^2)$ is an infinitely differentiable even function on $\Bbb{R}$.
Suppose that the claim holds for $k=1$, and letsuppose that $g^{(n)}(x)$ is defined for $x \ge 0$ and that $g^{(n)}(x^2)$ is even and infinitely differentiable. Then, we may define $h(x) = g^{(n)}(\sqrt{x})$, and from the base case we find that $h$ is differentiable and $h'(x^2)$ is even and infinitely differentiable. But $h(x) = g^{(n)}(x)$, so $g^{(n)}$ satisfies the required conditions. It follows that $g$ is infinitely differentiable, so long as we can prove that $g'(x)$ exists for $0 \le x$ and $g'(x^2)$ is infinitely differentiable. Let's try to prove that.
We have that for $x > 0$,
$$g'(x) = \frac{f'(\sqrt{x})}{2\sqrt{x}}$$
by the chain rule, and for $x=0$,
$$g'(0) = \lim_{h\to0} \frac{g(h) - g(0)}{h} = \lim_{h\to0} \frac{f(\sqrt{h}) - g(0)}{h}= \lim_{h\to0} \frac{f(0) + f'(0)\sqrt{h} + \frac{1}{2}f''(0)h + o(h) - f(0)}{h}$$
by Taylor's Theorem. Now, since $f$ is even, $f'(0) = 0$, so
$$g'(0) = \frac{1}{2}f''(0)$$
Define $h(x) = g'(x^2)$. Then,
$$h(x) = \begin{cases} \frac{f'(x)}{2x} & x \not= 0 \\ \frac{f''(0)}{2} & x = 0 \end{cases}$$
Since $f$ is even, we have at once that $f'(x)$ is odd, so $\frac{f'(x)}{2x}$ is even and $h$ is even.
I claim that $h$ is infinitely differentiable, with
$$h^{(k)}(x) = \begin{cases}\frac{1}{2x^{k}}\sum_{n=0}^k (-1)^{n+k} \frac{(k-1)!}{n!} x^n f^{(n+1)}(x) & x \not= 0\\ \frac{f^{(k+2)}(0)}{2(k+1)} & x=0\end{cases}$$
In the base case, the identity holds for $k=0$. Now, assuming the identity holds for $k$, we have that for $x\not=0$
$$h^{(k+1)}(x) = \frac{-k}{2x^{k+1}}\sum_{n=0}^k (-1)^{n+k} \frac{(k-1)!}{n!} x^n f^{(n+1)}(x) + \frac{x}{2x^{k+1}}\sum_{n=1}^k (-1)^{n+k} \frac{(k-1)!}{(n-1)!} x^{n-1} f^{(n+1)}(x)$$ $$+ \frac{x}{2x^{k+1}}\sum_{n=0}^k (-1)^{n+k} \frac{(k-1)!}{n!} x^n f^{(n+2)}(x)$$
$$= \frac{1}{2x^{k+1}}\sum_{n=0}^{k+1} (-1)^{n+k+1} \frac{k!}{n!} x^n f^{(n+1)}(x)$$
Since the terms $n=1,2,\ldots k$ in the second sum cancel with the terms $n=0,1,\ldots,k-1$ in the third sum.
And for $x=0$, we Taylor series expand each of the $f^{(m)}$ terms in $h^{(k)}$ to find that for $x \not=0$,
$$\begin{align*}
h^{(k)}(x) &= \frac{1}{2x^{k+1}} \sum_{n=0}^{k} (-1)^{n+k} \frac{k!}{n!} x^n f^{(n+1)}(x)\\
&= \frac{(-1)^k}{2x^{k+1}} \sum_{n=0}^k (-1)^n \frac{k!}{n!} x^n \sum_{m=n}{^k+2} \frac{f^{(m+1)}(0)x^{m-n}}{(m-n)!} + o(x^{k+2})\\
&= \frac{(-1)^k}{2x^{k+1}}\left(\sum{m=0}^{k+2} f^{(m+1)}(0)x^m \frac{k!}{m!} \sum_{n=0}^m \frac{(-1)^n m!}{n!(m-n)!}\right) + \frac{f^{(k+2)}(0)}{2(k+1)}
\\&\qquad+ \frac{xf^{(k+3)}(0)}{2}\left[\frac{1}{k+1} - \frac{1}{(k+1)(k+2)}\right] + o(x)\\
&= \frac{f^{(k+2)}(0)}{2(k+1)} + \frac{xf^{(k+3)}(0)}{2(k+2)} + o(x)
\end{align*}$$
Thus, $$h^{(k+1)}(0) = \lim_{x\to0}\frac{h^{(k)}(x) - h^{(k)}(0)}{x} = \frac{f^{(k+3)}(0)}{2(k+2)}$$
This establishes the desired result.
