If $n=1$, then
$$ f(x)= \begin{cases} 0 &~\text{for}~~ x \le 0 \\ e^{-\frac{1}{x^{2}}} &~\text{for}~~ x >0. \end{cases} $$ is infinitely differentiable but not analytic at the origin, because its Taylor series around origin does not equal $f(x)$ for $x >0$. Can I extend this function as follows:
$$ f(x_{1},x_{2},...,x_{n})= \begin{cases} 0 &~\text{for}~~ x_{1}x_{2}...x_{n} \ \le 0 \\ e^{-\frac{1}{x_{1}^{2}x_{2}^{2}...x_{n}^{2}}} &~\text{for}~~ x_{1}x_{2}...x_{n} >0. \end{cases} $$ But how I will check it is infinitely differentiable, because I need to check differentiability at some other points including origin. Please help me to figure out how we can extend $f$ from $\mathbb{R}$ to $\mathbb{R}^{n}$.
Take $f(x)=e^{-1/(\sum\limits_{k=1}^{n} x_k^{2})}$ for $x\neq 0$ and $0$ for $x=0$. It is clear that this functin is $C^{\infty}$ and not analytic