Let $f : \mathbb{R}^n \rightarrow \mathbb{R}^n$ be given by the equation $f(x)=\|x\|^2\cdot x$. Show that f is of class $C^{\infty}$ and that $f$ carries the unit ball $B(0,1)$ onto itself in a one-to-one fashion. Show, however, that the inverse function is not differentiable at $0$.
I already proved that $f$ is a class $C^{\infty}$ function and also I prove that sends to the unit ball $B(0,1)$ in a bijective way in itself, and arrives at the equal $f(B(0,1))=B(0,1)$ which proves that it is surjective, but I am stuck in trying to prove that it is injective, I suppose that $f(x)=f(z)$ taking into account the fact that $|x|<1$ and $|z|<1$ because $x\in B(0,1)$ and $z \in B(0,1)$ and get that $x=z$ I do not know how to follow, and prove that the inverse of f is not differentiable at 0
Injectivity: Suppose that $\|x\|^{2}x=\|y\|^{2}y$, then $\big\|\|x\|^{2}x\big\|=\big\|\|y\|^{2}y\big\|$, so $\|x\|^{3}=\|y\|^{3}$ and hence $\|x\|=\|y\|$, so from $\|x\|^{2}x=\|y\|^{2}y$ again we deduce that (assuming that $x\ne 0$) $x=y$.
One shows that $f^{-1}(y)=\dfrac{y}{\|y\|^{2/3}}$ for $y\ne 0$, and $f^{-1}(0)=0$. Assume that it is differentiable at $y=0$, then $\lim_{y\rightarrow 0}\dfrac{\|f^{-1}(y)-A(y)\|}{\|y\|}=0$ for some linear operator $A$. Realizing that $y=he_{1}$, $h>0$ and taking $h\rightarrow 0^{+}$, we are ready to get a contradiction.