how to prove a continuous, homogeneous map between two finite dimensional vector space is linear if it is differentiable at 0. By homogeneity, we can reduce the problem to a neighborhood of 0. Also, inverse function theorem tells me locally around 0, the continuous map is invertible. But I don't know how to proceed from here:
The vector spaces matters are $R^n$. But I am also interested to know if this proposition holds for general vector space and how to prove it.
And the definition of differentiablity is as follow: 
For $f: R^n\to R^m$ is differentiable at $a$  if there exists a linear transformation $R^n \to R^m$ such that $$\lim_{x\to a } \frac{|f(x)-f(a)-T(x-a)|}{|x-a|}$$ exists. 
The matrix representation of $T$ is denoted as $Df(a)$.
 A: Yes, it's true for arbitrary Real Banach spaces (complete normed vector spaces) as well. The proof is shown in Loomis and Sternberg's Advanced Calculus (page $148$, right after Theorem $7.2$)
I'll restate the argument here (but you should take a look at the example given there). The claim is:

Let $V$ and $W$ be Real Banach spaces, and let $f:V \to W$ be a homogeneous mapping of degree $1$; i.e for all $t\in \Bbb{R}$ and for all $\xi \in V$, we have $f(t\xi) = t f(\xi)$. Lastly, suppose also that $f$ is (Frechet) differentiable at the origin $0$. Then, $f$ is linear; In fact, $f = df_0$.
(I'm using $df_0$ to denote the linear map $T$ from $V$ to $W$ in your notation) 

To prove this, let $\xi \in V$ be arbitrary. We'll show that $f(\xi) = df_0(\xi)$. Define the map $\gamma: \Bbb{R} \to W$ by 
\begin{align}
\gamma(t) := f(t\xi) = t f(\xi).
\end{align}
Note that $\gamma(t) = tf(\xi)$ is a linear function of $t$ and hence is differentiable at the origin (actually it's everywhere differentiable) with
\begin{align}
\gamma'(0) = f(\xi)
\end{align}
Now, we can also differentiate $\gamma(t) = f(t\xi)$ using the chain rule, because $f$ is assumed to be differentiable at the origin (see Theorem $7.2$ if you need more clarification). Doing so yields:
\begin{align}
\gamma'(0) = df_{0}(\xi)
\end{align}
Hence, we have just shown that $f(\xi) = \gamma'(0) = df_0(\xi)$. Since $\xi \in V$ was chosen arbitrarily, it follows that
\begin{align}
f = df_0
\end{align}
as claimed, thereby proving that $f$ is linear.
(I hope you realize that the differentiability assumption on $f$ at the origin is important! We have to make us of it before applying the chain rule.)

We can actually generalize this statement slightly and claim the following:

Let $V,W$ be Real Banach spaces, and let $f:V \to W$ be a homogeneous function of degree $k$ (i.e for all $t \in \Bbb{R}$ and for all $\xi \in V$, $f(t\xi) = t^k f(\xi)$). If we assume in addition that $f$ is $k$ times (Frechet) differentiable at the origin, then $f$ is a homogeneous polynomial of degree $k$. In fact, for every $\xi \in V$, we have that
  \begin{align}
f(\xi) = \dfrac{1}{k!} d^kf_0(\underbrace{\xi, \dots, \xi}_{k \text{ times}})
\end{align}

This final formula in fact shows that $f$ is a degree $k$ polynomial and hence is $C^{\infty}$. For the proof of this, see my answer to this question.
