$f(x)=||x||^2 x$ is bijective on $\mathbb{R^n}$ Let $f: \mathbb{R^n} \to \mathbb{R^n}$ such that $f(x)= ||x||^2 x$. I want to show that $f$ is bijective.
We know local bijection using inverse function theorem at all points other than zero.Can we say a global inverse exists?

It is convenient to write your function as 
$$f(x) = ||x||^3 \cdot \frac{x}{||x||}$$
More generally, you can have 
$$f(x) = \phi(||x||) \cdot \frac{x}{||x||}$$
It is perhaps safe to assume extra that $\phi(0) = 0$. Now, if your function of $1$ variable $\phi\colon [0, \infty)\to [0, \infty)$ is bijective then $f$ is also bijective and the inverse will be 
$$f^{-1}(y) = \phi^{-1}(||y||) \cdot \frac{y}{||y||}$$
Check how it works in our example.
