$f(x,y) = f(αx,αy)$ for all $α>0$ and $(x,y)$ distinct from $(0,0)$
$f$ satisfies:
$$x\frac{\partial f}{\partial x} + y\frac{\partial f}{\partial y} = 0$$
My attempt:
what I tried is to do the implication go from 2) to 1). I divide by the partial derivative (I dont know if I can do this actually, I guessed than since the limit is just a value I could do it, and if that limit is equal to zero then going from 1) to 2) is trivial since $f$ is just a constant value) and obtain:
$$\frac{x}{-y} = \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}.$$
Then, using partial derivative definition, you get that $x=-y,$ which implies that the partial derivative respect to $x$ is equal to the partial derivative respect to $y.$
From this I tried to justify 1) but I just got stuck. I think I am really far from getting the question right but I wanted to know whether what I am doing right or wrong from my reasoning.