$f:\mathbb{R}^n\to\mathbb{R}^n,f(x)=x\|x\|^2$ Then $f:\mathbb{R}^n\to\mathbb{R}^n,f(x)=x\|x\|^2$ Then
$1.$ $(Df)(0)=0$
$2.$ $(Df)(x)=0\forall x$
$3.$ $f$ is one one
$4$. $f$ has an inverse.
let, $x=(x_1,x_2,x_3,\dots,x_n)$, $\|x\|^2=(x_1^2+\dots,+x_n^2)$
,Then $Df=3\|x\|^2$ am I right?  Then $1$ is true, $2,3,4$ are false.
 A: Your $f$ maps $(x_1,\dots,x_n)$ to $$((x\cdot x)x_1,\dots,(x\cdot x)x_n)$$
Let $f_k(x)=(x\cdot x)x_k$. Then $$\nabla f_k(x)=\nabla (x\cdot x)x_k+(x\cdot x)\nabla (x_k)$$
$$\nabla f_k(x)=2x_k x+(x\cdot x)e_{k}$$
Thus $$M_{(Df)(x)}=2 \begin{pmatrix} x_1x_1&x_1x_2&\cdots&x_1x_n\\x_2x_1&x_2x_2&\cdots&x_2x_n\\ \vdots &\vdots&\ddots&\vdots\\x_nx_1&x_nx_2&\cdots&x_nx_n\end{pmatrix}+\Vert x\Vert^2{\bf Id}$$
Note that your functions is of the form $f(x)=\lambda(x)\operatorname{id}(x)$ with $\lambda:\Bbb R^n\to \Bbb R$. We know that $\lambda(x)$ is not injective, but we know that $\operatorname{id}(x)$ is. Note $f$ is injective on every $\{\Vert x\Vert =r\}$, so we must look at $x,y$ with $\Vert x\Vert \neq \Vert y \Vert$, assume $\Vert x\Vert \geq \Vert y \Vert$. Since the norm is always non-negative, $f$ may only fail to be injective when each coordinate of $x$ has the same sign as the corresponding coordinate of $y$. Could you look at what happens in said case? Aim to show $x\neq y\implies f(x)\neq f(y)$.
A: No, if you are looking at the derivative of a map between two vector spaces, it has to be a linear map. In this context, the derivative of $f$ at $x$ is an endomorphism $Df(x):\mathbb R^n\rightarrow \mathbb R^n$.
So when you give $Df=3\|x\|^2$, it can't be a good answer for $n\geq 2$. However if $n=1$, I agree.
To answer question 1) and 2), you can easily compute $Df(x)$ for any $x\in \mathbb R^n$ : since $f(x)=x\|x\|^2=\langle x,x\rangle x$, $$f(x+h)-f(x)=\langle x+h,x+h\rangle (x+h)-\langle x,x\rangle x= \cdots$$
For question 3), assume that $f(x)=f(y)$ and try to prove that $x=y$ i.e. $f$ is injective.
Then by question 3), if $f$ is injective, what else does $f$ need to be invertible ?
