Prove that for a homogeneous function of degree one all directional derivatives exist I am trying to prove that for a function $f:\mathbb{R}^n \rightarrow\mathbb{R}$ that is homogeneous of degree one, all directional derivatives exist. I also want to prove that it is differentiable if and only if $f$ is linear. 

I know that since the function is homogeneous, I have $f(\lambda x)=\lambda f(x)$ for $\lambda&gt;0$ and $x$ in $\mathbb{R}^n$, but don't know how to continue from this.
Let $0\ne (a,b)=h\in\mathbb{R}^2$ a direction. $f(0,0)=0\cdot f(1,0)=0$.

$(Df)(0,0)(h)=\lim_{t\rightarrow 0} \frac{f((0,0)+t(a,b))-f(0,0)}{\left \| (a,b) \right \|} =\lim_{t\rightarrow 0}\frac{f(ta,tb)}{t\cdot\sqrt{a^2+b^2}} =\lim_{t\rightarrow 0}\frac{t\cdot f(a,b)}{t\sqrt{a^2+b^2}}=\lim_{t\rightarrow 0}\frac{f(a,b)}{\sqrt{a^2+b^2}}=\frac{f(a,b)}{\sqrt{a^2+b^2}}$.
Let $0\ne (a,b)=h\in\mathbb{R}^2$ a direction. $f(0,0)=0\cdot f(1,0)=0$.

$(Df)(0,0)(h)=\lim_{t\rightarrow 0} \frac{f((0,0)+t(a,b))-f(0,0)}{\left \| (a,b) \right \|} =\lim_{t\rightarrow 0}\frac{f(ta,tb)}{t\cdot\sqrt{a^2+b^2}} =\lim_{t\rightarrow 0}\frac{t\cdot f(a,b)}{t\sqrt{a^2+b^2}}=\lim_{t\rightarrow 0}\frac{f(a,b)}{\sqrt{a^2+b^2}}=\frac{f(a,b)}{\sqrt{a^2+b^2}}$.