Linearity of $D_{v}f(0)$ 
Exercise: Let $f:\Bbb{R^n}\rightarrow \Bbb{R}$ be homegeneous of degree 1, in the sense that $f(tx)=tf(x)$ for all $x\in\Bbb{R^n}$ and $t\in\Bbb{R}$. Show that $f$ has directional derivatives at 0 in all directions. Prove that $f$ is differentiable at 0 if and only if $f$ in linear.
Thm 1: Let $f$ be as in Definition 2.3.1 ($f: U\rightarrow R^n$, where $U \subseteq R^n$ is an open set and $a \in U$) . If $f$ is differentiable at $a$ it has the following property: $f$ has directional derivatives at $a$ in all directions $v \in R^n$ and $D_v f(a) =
Df(a)v$. In particular, the mapping $R^n \rightarrow R^p$ given by $v \rightarrow D_v f (a)$ is
  linear.

Let $f$ be homogeneous of degree 1. We see that :
$$D_vf(0)=\lim_{t\rightarrow 0}{\frac{f(0+tv)-f(0)}{t}}=f(v)$$
So we can conclude that $f$ has directional derivative at 0 in all directions. Assume now that $f$ is differentiable at 0. By using Thm 1 we can say that $v\mapsto D_vf(0)=f(v)$ is linear. But the other way is where I got stuck? Any hints about how to prove this?

Every linear function on $\mathbb{R}^n$ is continuous. Therefore $f(x)=\langle a,x \rangle$ for a suitable $a \in \mathbb{R}^n$. Now it is a simple exercise to prove that $Df(0)=a$.
