I have tried an attempt on the following:

Let $F:U\in \Bbb{R}^n\to \Bbb{R}$ and $f:\Bbb{R}\to \Bbb{R}$ where $f$ is an even function. Now $F(\mathbf{x})=f(|\mathbf{x}|)$, where $|\ . |$ is the Euclidean norm.
Given $f$ is $C^r$, I have to show that $F$ is $C^r$.

My attempt was to argument by induction. I showed that $F$ is $C^1$, basically using univariate chain rule on $\frac{\partial f\left(\sqrt{x\_1^2+\dots+x\_n^2}\right)}{\partial x\_1}$ and arguing that it would be continuous using the fact that $f$ is $C^r$. Since the argument works for all $x\_i$, it establishes first-differntiability for $F$ and continuity of the operator $DF(\mathbf{x})$. Then I moved on to show $F$ is $C^r$ assuming $F$ is $C^{r-1}$ and $f$ is $C^r$. Now what I did was partially differentiate:

$$
{\partial \over \partial x\_2} \left[\frac{\partial ^{k-1}f}{\partial x\_1^{k-1}}\right] = \underbrace{{\partial \over \partial |\mathbf{x}|} \left[\frac{\partial ^{k-1}f}{\partial x\_1^{k-1}}\right]}\_{(1)} . \underbrace{{\partial |\mathbf{x}| \over \partial x\_2}}\_{(2)}
$$

My argument runs as follows: this partial derivative is continuous since $f$ is $C^r$ and hence $(1)$ is continuous. Continuity of $(2)$ is establishd in a similar manner, since $\sqrt{\mathbf{x}}$ is $C^r$ if $\mathbf{x} \neq \mathbf{0}$. Then the product would be continuous. I am uneasy on multivariate arguments and dont know if my approach is completely off track. I didnt use the even function property, so I am sure I might me missing something. Any hints and references would be welcome. Thanks!