I'd like to know if the function $ f(x) = e^{||x||_2^2}$ with $x \in \mathbb{R}^n $, is infinitely differentiable. Here's my thoughts on it.
For two variables,
$f(x) = e^{||x||_2^2} = e^{x_1^2 + x_2^2}$
so $\frac{\partial}{\partial x_1} f(x) = 2 x_1 e^{x_1^2} e^{x_2^2}$, $\frac{\partial}{\partial x_2} f(x) = 2 x_2 e^{x_1^2} e^{x_2^2}$ so $df = 2 x_1 e^{x_1^2} e^{x_2^2} + 2 x_2 e^{x_1^2}e^{x_2^2}$ which is clearly continuous as the product and sum of continuous functions is continuous. It can be written as $df = P(x)f(x)$ with $P(x) = 2 x_1 + 2x_2$ a polynomial function of two variables. I have the intuition that for any number of variable and any number of derivation we can always write
$$ d^n f(x) = P(x) f(x)$$
Is my intuition correct ? In this case as as polynomial function is differentiable on $\mathbb{R}^n$ and $f$ is also, the product of the two is also differentiable on $\mathbb{R}^n$ and so It would give me the answer. Is this reasoning correct ?
Your intuition is correct.
For a thorough proof, the best way would be to use the fact that the composite of two smooth functions is also a smooth function.