Smoothness at the origin of a radial function obtained by rotating an even function

First, I will show that the function $h:\mathbb{R}\to\mathbb{R}$ defined by $$ h(x) = \begin{cases} \frac{f'(x)}{x} & \text{if $x\neq 0$} \\ f''(0) & \text{if $x=0$}\end{cases} $$ is even and smooth. By L'Hôpital's rule, $h$ is continuous at $x=0$. The evenness of $h$ follows from the fact that $f'$ is odd. To show the smoothness of $h$, we can calculate that for $x\neq 0$ and $m\geq 1$, \begin{align*} h^{(m)}(x) &= \sum_{k=0}^m \binom{m}{k} f^{(m-k+1)}(x)\cdot(-1)^kk!x^{-k-1} \\ &= \frac{\sum_{k=0}^m (-1)^k k!\binom{m}{k}x^{m-k}f^{(m-k+1)}(x)}{x^{m+1}}. \end{align*} We use L'Hôpital's rule to find $\lim_{x\to 0} h^{(m)}(x)$. The limit is of the indeterminant form $\frac{0}{0}$, and we get \begin{align} \lim_{x\to 0} h^{(m)}(x) &= \lim_{x\to 0} \frac{\sum_{k=0}^{m-1} (-1)^kk!\binom{m}{k}(m-k)x^{m-k-1}f^{(m-k+1)}(x) + \sum_{k=-1}^{m-1}(-1)^{k+1}(k+1)!\binom{m}{k+1}x^{m-k-1}f^{(m+1-k)}(x)}{(m+1)x^m} \\ &= \lim_{x\to 0} \frac{x^mf^{(m+2)}(x)}{(m+1)x^m} \\ &= \frac{f^{(m+2)}(0)}{m+1}. \end{align} If we assume that $h^{(m-1)}(x)$ is continuous, then the above shows (see Rudin's Principles of Mathematical Analysis, Exercise 5.9) that $h^{(m)}(0)$ exists and moreover that $h^{(m)}(x)$ is continuous at $x=0$. By induction, $h$ has derivatives of all orders.

Now back to the original problem. I noted in my statement of the question that we can calculate that $$ \frac{\partial g}{\partial x^i} = \begin{cases} h(|x|)x^i & \text{if $x\neq 0$} \\ 0 & \text{if $x=0$.}\end{cases} $$ This shows that $g$ is $C^1$. We have thus proven for $k=1$ the statement "Given any $f:\mathbb{R}\to\mathbb{R}$ smooth and even, $g:\mathbb{R}^n\to\mathbb{R}$ defined by $g(x)=f(|x|)$ is $C^k$". Assuming the statement proved for $k$, it is also proved for $k+1$. For the above expression shows that each partial derivative of $g$ is $C^k$ (we apply the induction hypothesis to $h$). This completes the proof.