which of the following is/are correct? 
Let $f:\mathbb R^m \to \mathbb R^m$ be a differentiable  function. let
  $Df(x)$ be the derivative of $f$ at $x\in \mathbb R^m$. which of the
  following is/are correct?
(a)$Df(0)(u)=0, \forall u \in \mathbb R^m$
(b)$Df(x)(u)=0, \forall u \in \mathbb R^m$ and some $x\in \mathbb R^m$
  only if $f$ is a constant.
(c)$Df(x)(u)=0, \forall u \in \mathbb R^m$ and all $x\in \mathbb R^m$
  only if $f$ is a constant.
(d)If $f$ is not a constant function, then $Df(x)$ is a one-one
  function for some  $x\in \mathbb R^m$.

(a) Need not true. since at $0\in \mathbb R^m$, $Df(0)=O$(zero transformation). analogous to derivative in real variable. it may have non zero linear transformation as a derivative at $0$. But I couldn't find the counterexample.
(b) I am not able to find the counter example.
(c)True, analogous to the result $f'(x)=0$ in a connected domain, then $f(x)=c$. but I am not able to prove the generalized result. I know that $\mathbb R^m$ is connected.
(d) If $f$ is not a constant function, then $Df(x)$ is a linear transformation. How to prove that it is one-one? for finite dimensional transformation one-one and onto are equivalent. Please help me
 A: a.)  Take $f(x) = Ax$ for some $m \times m$ matrix $A \ne 0$; then $Df(x)(u) = Au$ for all $x \in \Bbb R^m$, i.e. $Df(x) = A \ne 0$.  So $Df(0)(u) \ne 0$ for some $u$;
b.)  Consider the function $f:\Bbb R^m \to \Bbb R^m$ defined by
$f(x) = \begin{pmatrix} \sum_1^m x_i^2 \\ \sum_1^m x_i^2 \\ \vdots \\ \sum_1^m x_i^2 \end{pmatrix}, \tag 1$
where 
$x = (x_1, x_2, \ldots, x_m) \in \Bbb R^m; \tag 2$
that is, each component of $f(x)$ is the same function $\sum_1^m x_i^2$ of $x$.  Then
$Df(x) = \begin{bmatrix} 2x_1 & 2x_2 & \ldots & 2x_m \\  2x_1 & 2x_2 & \ldots & 2x_m \\ \vdots \\  2x_1 & 2x_2 & \ldots & 2x_m \end{bmatrix}; \tag 3$
that is, every row of $Df(x)$ is the same vector $(2x_1, 2x_2, \ldots, 2x_m)$; then
$Df(0) = 0, \tag 4$
so 
$Df(0)(u) = 0, \; \forall u \in \Bbb R^m, \tag 5$
but $f(x)$ is not constant;
c.)  here we use integration; for $x \in \Bbb R^m$ consider the path
$a(t) = tx, \; 0 \le t \le 1; \tag 6$
then
$f(x) - f(0) = \displaystyle \int_0^1 \dfrac{df(a(s))}{ds}ds = \int_0^1 Df(a(s)) \left (\dfrac{da(s)}{ds} \right ) ds = \int_0^1 0 \; ds = 0, \tag 7$
since
$Df(a(s)) = 0; \tag 8$
thus, for all $x \in \Bbb R^m$,
$f(x) = f(0), \tag 9$
and $f(x)$ is constant.
d.)  take 
$f(x) = Ax, \tag{10}$
where $A \ne 0$ is singular $m \times m$ matrix.  Then $f(x) \ne 0$ but
$Df(x) = A \tag{11}$
is nowhere one-to-one, since $\ker A \ne \{0\}$.
In short, only item (c) is true; (a), (b) and (d) are false.
