$k(tx,ty)=tk(x,y)$ 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 ? 

In your class did you do the theorem that if $f(x,y)$ is differentiable at $(x_0,y_0)$ then its directional derivative exists there for every direction $v$, given by $\partial_v f(x_0,y_0) = \nabla f(x_0,y_0) \cdot v$? If so, it's very straightforward. Take $(x_0,y_0) = (0,0)$. Then you have
$$\nabla f(0,0) \cdot (x,y) = \partial_{(x,y)} f(0,0)$$
$$= \lim_{t \rightarrow 0} {f(tx,ty) - f(0,0) \over t}$$
$$= \lim_{t \rightarrow 0} {f(tx,ty) \over t}$$
$$= f(x,y)$$
So you have $f(x,y) = \nabla f(0,0) \cdot (x,y) = Ax + By$, where $A = {\partial f \over \partial x}(0,0)$ and $B = {\partial f \over \partial y}(0,0)$.
