show that

$\displaystyle f(x,y,z)=\frac{1}{z^2+1}(x^2-y^2,2xy,2z)$ defines a smooth function $\displaystyle f: S^{2}\to S^{2}$ where $\displaystyle S^{2}=\{p\in \mathbb{R}^3, |p|=1\}$ so I am guessing it is a unit sphere.

How do we proceed with such problem?

f is a smooth function if all partial derivatives of all possible orders are defined in all points of the domain of f right?

I started with computing these partial derivatives

$\displaystyle \frac{\partial f}{\partial x}=\frac{1}{z^2+1}(2x,2y,0)$

$\displaystyle \frac{\partial f}{\partial y}=\frac{1}{z^2+1}(-2y,2x,0)$

$\displaystyle \frac{\partial f}{\partial z}=\frac{-2(z^2-1)}{(z^2+1)^2}$

and these partial derivatives are all defined at any point from the domain of f

so I compute second order partial derivatives

$\displaystyle \frac{\partial^2 f}{\partial x^2}=\frac{2}{z^2+1}$

$\displaystyle \frac{\partial^2 f}{\partial y^2}=\frac{-2}{z^2+1}$

$\displaystyle \frac{\partial^2 f}{\partial z^2}=\frac{4z(z^2-3)}{(z^2+1)^3}$

and these 3 are also defined for all the points of the domain.

if I continue differentiating

$\displaystyle \frac{\partial^3 f}{\partial x^3}=\frac{\partial^3 f}{\partial y^3}=0$ and

$\displaystyle \frac{\partial^3 f}{\partial z^3}=-\frac{12(z^4-6z^2+1)}{(z^2+1)^4}$ which is defined for all points

What should I do next? can I conclude based on the calculations that f is certainly smooth?

Thanks in advance