$k(tx,ty)=tk(x,y)$ 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 ? 
 A: Immediately, we have $k(0,0)=0$. Put $A=k_x(0,0)$ and $B=k_y(0,0).$ Then $k(0,0)=0=A\cdot 0+B\cdot 0,$ and for any non-zero vector $(u,v),$ we have $$\begin{align}Au+Bv &= uk_x(0,0)+vk_y(0,0)\\ &= \sqrt{u^2+v^2}\cdot\frac{uk_x(0,0)+vk_y(0,0)}{\sqrt{u^2+v^2}}\\ &= \sqrt{u^2+v^2}\left(\nabla_{(u,v)}k(0,0)\right)\\ &= \sqrt{u^2+v^2}\left(\lim_{t\to 0}\frac{k(tu,tv)-k(0,0)}{t\sqrt{u^2+v^2}}\right)\\ &= \sqrt{u^2+v^2}\left(\lim_{t\to 0}\frac{tk(u,v)}{t\sqrt{u^2+v^2}}\right)\\ &= \sqrt{u^2+v^2}\cdot\frac{k(u,v)}{\sqrt{u^2+v^2}}\\ &= k(u,v),\end{align}$$ where $\nabla_{(u,v)}k(0,0)$ denotes the directional derivative of $k$ at $(0,0)$ in the direction of $(u,v)$. Thus, $k(x,y)=Ax+By$, as desired.
A: Take the derivative with respect to $t$ of each side of $k(tx,ty)=tk(x,y)$, using the chain rule on the left side, while on the right the $t$ drops out, obtaining
$$\frac{\partial k}{\partial x}(tx,ty)\cdot x + \frac{\partial k}{\partial y}(tx,ty)\cdot y =k(x,y).$$
Now put $t=0$ and obtain
$$\frac{\partial k}{\partial x}(0,0)\cdot x + \frac{\partial k}{\partial y}(0,0)\cdot y =k(x,y).$$
This is your expression with $A,B$ taken as the two partials at $(0,0).$
A remark about notation: Given any function of two variables, say $g(u,v)$, it is better to use $g_1(u,v)$ and $g_2(u,v)$ for the two partials w.r.t. $u$ and $v$ in the context of the chain rule. Each partial is a function of two variables. If then $u=a(t),\ v=b(t)$ the function $g(a(t),b(t))$ is a function of the one variable $t$ and the chain rule says that
$$\frac{d}{dt}g(a(t),b(t))=g_1(a(t),b(t))\cdot a'(t)+g_2(a(t),b(t))\cdot b'(t).$$
In the case above, we take $g=k$ and $a(t)=tx,\ b(t)=ty$, obtaining for the chain rule side of the derivative of the equation w.r.t. the variable $t$ as the left side of
$$k_1(tx,ty)\cdot x + k_2(tx,ty)\cdot y=k(x,y),$$
while the right side is clearly $[tk(x,y)]'=k(x,y).$
At this point $t$ may be set to $0$ as above in the (more confusing) usual partial derivative notation.
A: First $k(0,0)=0$. $k(x, y)=\lim_{t\to 0}\frac{k(tx, ty)}{t}=xk_x+yk_y$ where $k_x=\partial_xk|_{(x,y)=(0,0)}, k_y=\partial_yk|_{(x,y)=(0,0)}$.
A: 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 $B = {\partial f \over \partial y}(0,0)$.
