Show that $f:\Bbb R^m\to\Bbb R,\: x\mapsto [|x|<1] \exp\left(\frac1{|x|^2-1}\right)$ is smooth, where $[|x|<1]$ is a Iverson bracket.
My work so far: $f|_{\{x\in\Bbb R^m:|x|<1\}}$ is a composition of smooth functions, hence by the chain rule it can be shown that it is also smooth, and $f|_{\{x\in\Bbb R^m:|x|\ge 1\}}=0$ is trivially smooth. Then we must check that $f$ is differentiable in $\mathrm S^{m-1}$, that is, we must show that
$$\lim_{h\to 0}\frac{\|\partial^k f(x+h)-\partial^k f(x)-\partial^{k+1}(x)h\|}{\|h\|}=0,\quad\forall k\in\Bbb N_{>0},\:\forall x\in\mathrm S^{m-1}\tag1$$
I guess that $\partial^k f(x)=0$ for all $x\in\mathrm S^{m-1}$ and for all $k\in\Bbb N$. Then, assuming this hypothesis, from $(1)$ the proof was reduced to show that
$$\partial^k f(x+h)\in o(\|h\|)\quad\text{for }|x+h|<1,\:x\in\mathrm S^{m-1}\tag2$$
Because $f$ is smooth in this region using the Taylor theorem we have that
$$f(x+h)=\sum_{k=0}^n\frac{\partial^kf(h)[x]^k}{k!}+o(\|x\|^n),\quad\forall n\in\Bbb N\tag3$$
where $o(\|x\|^n)=o(1)$ because $x\in\mathrm S^{m-1}$. However I cant stablish a clear relation between $(2)$ and $(3)$ to end the proof. I need some help with this exercise.
Prove the following Lemma using induction.
Lemma. The function $q(t):=e^{-1/t}$ $(t>0)$ and $:=0$ $(t\leq0)$ is smooth, whereby $q^{(n)}(0)=0$ for all $n\geq0$.
Your $f$ is just $q\circ g$, where $g(x):=1-|x|^2$. The chain rule then guarantees the claim in your question. When required do an additional induction proof for this. But it is not necessary to compute partial derivatives.