$k(tx,ty)=tk(x,y)$ then $k(x,y)=Ax+By$ A friend asked me today the following question:

Let $k(x,y)$ be differentiable in all $\mathbb{R}^{2}$ s.t for every
  $(x,y)$ and for every $t$ it holds that $$k(tx,ty)=tk(x,y)$$ Prove that
  there exist $A,B\in\mathbb{R}$ s.t $$k(x,y)=Ax+By$$

I want to use the chain rule somehow, but I am having difficulty using
it (I am a bit rusty). I believe I can get $$\frac{\partial k}{\partial tx}\cdot\frac{\partial tx}{\partial t}+\frac{\partial k}{\partial ty}\cdot\frac{\partial ty}{\partial t}=k(x,y)$$
hence $$\frac{\partial k}{\partial tx}\cdot x+\frac{\partial k}{\partial ty}\cdot y=k(x,y)$$
but I don't see how this helps.
Can someone please help me out ? 

First $k(0,0)=0$. $k(x, y)=\lim_{t\to 0}\frac{k(tx, ty)}{t}=xk_x+yk_y$ where $k_x=\partial_xk|_{(x,y)=(0,0)}, k_y=\partial_yk|_{(x,y)=(0,0)}$.
