Let $f:\Bbb R^{n} \to \Bbb R^{n}$ be the function defined by $f(x)=x\left \|x \right \|^{2}$ for $x \in \Bbb R^{n}$. I came across the problem which says:
Let $\displaystyle f:\Bbb R^{n}\rightarrow \Bbb R^{n}$ be the function defined by $f(x)=x\left \|x  \right \|^{2}$ for $x \in \Bbb R^{n}$. Then which of the following statements are correct?
(a) $(Df)(0)=0$,
(b) $(Df)(x)=0$ for all $x$ belongs to $\Bbb R^{n}$,
(c) $f$ is one-one,
(d) $f$ has an inverse.
I do not know how to approach the problem.Any kind of hints will be helpful. Thanks everyone in advance for your time .
 A: $\def\R{\mathbb R}\def\norm#1{\left\|#1\right\|}\def\abs#1{\left|#1\right|}\def\sp#1{\left\langle#1\right\rangle}$Recall that for a function $F\colon \R^n \to \R^n$ to be differentiable at $x \in \R^n$ there must exist a linear $DF(x) \colon \R^n \to \R^n$ such that $$ F(x+h) = F(x) + DF(x)h + o(h), \qquad h \to 0 $$ For the given $F$ we have for $x,h\in \R^n$
\begin{align*}
  F(x+h) - F(x) &= \norm{x+h}^2(x+h) - \norm x^2x\\
         &= \sp{x+h,x+h}(x+h) - \norm x^2x\\
         &= \bigl(\norm x^2 + 2\sp{x,h} + \norm h^2\bigr)(x+h) - \norm x^2 x\\
         &= \norm x^2h + 2\sp{x,h}x + \norm h^2x + 2\sp{x,h}h + \norm h^2 h
\end{align*}
Note that $DF(x)h := \norm x^2 h + 2\sp{x,h}x$ is linear in $h$ and 
\begin{align*}
  \norm{\norm h^2x + 2\sp{x,h}h + \norm h^2 h} &\le \norm h^2\norm x + 2\norm x\norm h^2
   + \norm h^3\\
   &= \norm h \bigl(2\norm h\norm x + \norm h^2\bigr)\\
   &= o(\norm h), \qquad h \to 0
\end{align*}
So we found the derivative. So (i) is true as for any $h$ 
$$ DF(0)h = \norm 0^2 h + 2\sp{0,h}0 = 0 $$
but (ii) is wrong.
For (iii) and (iv) we observe that for $x \ne 0$
$$ \norm{F(x)} = \norm{x\norm x^2} = \norm x^3 \iff \norm x^2 = \norm{F(x)}^{2/3} \iff x = \frac{F(x)}{\norm{F(x)}^{2/3}} $$
Hence if we define $G \colon \R^n \to \R^n$ by 
$$ G(x) = \begin{cases} 0 & x = 0\\[3mm] \frac{x}{\norm x^{2/3}} & \text{otherwise} \end{cases} $$
the above shows that for $x \ne 0$ we have $x = G\bigl(F(x)\bigr)$ and as $F(0) = 0$ we moreover have $(G\circ F)(0) = 0$. On the otherside, $(F \circ G)(0) = 0$ and for $x \ne 0$ we have
$$ F \bigl(G(x)\bigr) = \norm{G(x)}^2G(x) = \norm{\frac{x}{\norm x^{2/3}}}^2\frac{x}{\norm{x}^{2/3}} = \frac{\norm x^2}{\norm x^{6/3}} x = x $$
So $F \circ G = G \circ F = \mathrm{id}_{\R^n}$ and $G$ is $F$'s inverse. Hence (iii) and (iv) are true.
