I was solving a problem which boiled down to checking the smoothness of the following function.
$$ \tilde{p} (z) = \begin{cases} \frac{1}{p(1/z)} &\quad z\neq 0 \\ 0 &\quad z = 0.\end{cases} $$
where $p$ is a polynomial not identically zero and the domain of the function is chosen so that $p(1/z) \neq 0$ in all of the domain.
Continuity of the function is easy to see as $z \to 0$ implies $p(1/z) \to \infty$ hence $\tilde{p}(z) \to 0$. But I do not see how to imply smoothness of $\tilde{p}(z)$. Here I do not want it to be holomorphic, it is enough if I can see that $\tilde{p}$ is smooth as function of two variables.
The reciprocal map $\rho: \ z\mapsto{1\over z}$ is continuous througout its domain of definition $\dot{\mathbb C}:={\mathbb C}\setminus\{0\}$. Let $\Omega$ be the envisaged domain of $\tilde p$. Since it is assumed that $p(1/z)\ne0$ for all points $z\in\dot \Omega$ the composition $$\tilde p(z)=(\rho\circ p\circ\rho)(z)\qquad(z\in\dot\Omega)$$ is continuous throughout $\dot\Omega$.
The point $z=0$ needs special treatment. If ${\rm deg}(p)\geq1$ then $$\lim_{z\to0}p(1/ z)=\lim_{w\to\infty}p(w)=\infty\ ,$$ and therefore $$\lim_{z\to0}\tilde p(z)=\lim_{z\to0}{1\over p(1/z)}=0=\tilde p(0)\ .$$ It follows that $\tilde p$ is continuous at $z=0$ in this case. If, however, ${\rm deg}(p)=0$ then $p(z)\equiv c\ne0$, and we have $$\lim_{z\to0}p(1/ z)=\lim_{w\to\infty}p(w)=c\ ,$$ and therefore $$\lim_{z\to0}\tilde p(z)=\lim_{z\to0}{1\over p(1/z)}={1\over c}\ne\tilde p(0)\ ,$$ so that $\tilde p$ is not continuous at $0$ in this case.