If $f(\alpha x, \alpha y) = f(x,y)$ is $f$ some special function? Like the title reads, if $f(\alpha x, \alpha y) = f(x,y)$ is $f$ some special function? Assume $x,y,\alpha\in\mathbb{R}$.
 A: The point $(0,0)$ is going to be a nuisance to you in two ways.
One, if you allow $\alpha = 0$ in your equation, then for all $x, y$, we have $f(x,y) = f(\alpha x, \alpha y) = f(0,0)$, so your function is constant.
Okay, fine, so let's say that $\alpha \not= 0$, or $\alpha > 0$. But now even if you want $f$ to just be continuous at $(0,0)$, you've still got a problem, because we should then have \[f(0,0) = \lim_{\alpha \to 0} f(\alpha x, \alpha y) = f(x, y) \] and the function is constant again.
So in order for $f$ to be non-constant, you need to either leave it undefined at $(0,0)$ or give it some value there that's essentially unconnected to anything else going on. Once you've done that, you can exmaine how the function behaves everywhere else: it's essentially a function that is constant along lines going through the origin. 


*

*If you stick with $\alpha \not= 0$, so that e.g. $f(x,y) = f(-x,-y)$, then as another answer has discussed you basically have a function on the real projective line, a function that factors through the quotient map $\mathbb R^2 \setminus \{(0,0)\} \to \mathbb{RP}^1$.

*If you decided to use $\alpha > 0$ instead, then your function would be constant on every ray emanating from the origin. You can assign an angle to each ray, and define your function as a function of that angle. One way to think about this is considering $(x,y)$ as a point on the complex plane, and your $f$ as a function of the number's argument (or phase), $f(x,y) = g(\arg(x + iy))$. Another way is to think about it as a function defined on a circle around the origin, or a function defined on the set of "directions from the origin". You factor through the quotient map $\mathbb {R}^2 \setminus \{(0,0)\} \to S^1$, the circle, given by dividing each vector by its length.


If you're unfamiliar with the terminology "$f$ factors through $g$", just let me know and I'll explain that in a bit more detail.
A: What comes to mind for me is the map which takes points in $\mathbb R^2$ and associates them with lines through the origin - which are the points of real projective space $\mathbb {RP}$. Points in real projective space are identified by coordinates similarly to that in $\mathbb R$, but one less coordinate is required. In this case, lines can be specified by a point they go through (since the origin is implied). However, scalar multiples of such points are the same line!
Here are some links so you can study this topic a little further.
https://en.wikipedia.org/wiki/Real_projective_space
http://mathworld.wolfram.com/RealProjectivePlane.html
A: Functions defined on $\dot{\mathbb R}^n:={\mathbb R}^n\setminus\{{\bf 0}\}$ and satisfying an identity of the form
$$f(\lambda{\bf x})=\lambda^d\>f({\bf x})\qquad\bigl(\lambda\in\dot{\mathbb R}, \ {\bf x}\in\dot{\mathbb R}^n\bigr)$$
(or $=|\lambda|^d\>f({\bf x})$, depending on the context) for some constant $d\in{\mathbb R}$ are called homogeneous of degree $d$. Such functions are completely determined by their values on $S^{n-1}$. Your function $f$ is homogeneous of degree $0$.
A: Here's a hopefully-straightforward way of seeing how other folks' solutions come about..  For simplicity I'll eliminate the 'awkward' cases where any of $x,y,\alpha$ are zero or negative, and consider just $x, y, \alpha\gt0$.
We can start by executing a simple change of variables: let $z=\frac xy$ and $w=xy$.  It's easy to see that we don't lose any data by doing this, since we can invert these relations to get $x=\sqrt{zw}$ and $y=\sqrt{\frac wz}$ (remember we're limiting our variables to being positive here).  Then we can define a function $g(z,w)=f(x,y)$; since $z(\alpha x,\alpha y)=\frac{\alpha x}{\alpha y}=\frac xy=z(x,y)$ and $w(\alpha x, \alpha y)={\alpha x}{\alpha y}=\alpha^2xy=\alpha^2w(x,y)$, then your relation can be written as $g(z,w)=g(z,\alpha^2 w)$.  But by choosing the appropriate value of $\alpha$ (namely, $\alpha=\sqrt{\dfrac{w'}w}$) this is the same as saying that $g(z,w)=g(z,w')$ for all $z,w,w'$.  In other words, $g(z,w)=g(z)$ is actually a function of one variable — or translating back, $f(x,y)$ is actually only a function of $\frac{x}{y}$.  This is equivalent to Alfred's comment that the function is essentially defined on the projective line.
Here's an example: suppose that we have $f(x,y)=\frac{x^2-y^2}{(x-y)^2}$; it's obvious from quick computation that $f(\alpha x,\alpha y)=f(x,y)$.  Now, let's expand $(x-y)^2=x^2-2xy+y^2$ in the denominator and then plug in $x=\sqrt{zw}$ and $y=\sqrt{\frac wz}$.  We get $g(z,w)=\dfrac{zw-\frac wz}{zw-2w+\frac wz}$.  This isn't immediately obvious as a function of $z$ alone, but notice that $w$ is a factor of both the numerator and denominator; on dividing it out and multiplying by $z$ on top and bottom (in other words, multiplying numerator and denominator both by $\frac zw$) we get $g(z,w)=\frac{z^2-1}{z^2-2z+1}=\frac{z^2-1}{(z-1)^2}$ (which just equals $\frac{z+1}{z-1}$ everywhere it's defined).  Alternately, we can 'take it on faith' that $g(z,w)$ doesn't depend on $w$ and say $g(z)=g(z,1)=\dfrac{z-\frac1z}{z-2+\frac1z}$, then multiply numerator and denominator by $z$ to clear the fractions, getting the same result.  This example also shows how the trickiness around zero can come in when doing the algebraic manipulations; note that $w=0$ if either $x=0$ or $y=0$, and so dividing out by it isn't necessarily allowed — even though the original function was well-defined everywhere away from the line $x=y$.
