Let $f$ be a real-valued function on $\mathbb R^3$ satisfying.... 
I came across the above problem and have to determine which of the above options is correct. Can someone point me in the right direction?Thanks in advance for your time.
EDIT:
If I take $x=r \frac {x}{||x||}$ and $y=r \frac {y}{||y||}$ as @uncookedfalcon suggested,then I see that $f(x)=f(rx)$ and  $f(y)=f(ry)$ since $||x||=||y||=1$ and hence  by using the fact that $f(x)=f(y)$ we see ,$f(rx)=f(ry)$ and thus $r^{\alpha}[f(x)-f(y)]=0.$ Also since $r=||x||=1 \neq 0$,so $f(x)=f(y)$=CONSTANT...
 A: Well, 1 implies 2 (take $\beta$ = 1), so probably not 1. To see not 4, the function $\| x \|^\alpha $ satisfies this property, and sure isn't constant. Between 2 and 3, 2 implies 3 ($c = 1$), so it can't be 2. So, try proving 3 (hint: write any $x$ as $r\frac{x}{\| x\|}$, where $r = \| x \|$). 
A: To finish off the question, pick any $x$ with $\|x\|=1$, and define $c$ by $f(x)=c$. Now given any $y$, write $y=\|y\|v$ where $\|v\|=1$, and calculate $f(y)=f(\|y\|v)$ using the properties given for $f$. 
EDIT: the discussion in the comments suggests amplification is needed, so here goes. 
We are assuming (#) there is a real number $a$ such that for all positive $r$ and all vectors $x$ we have $f(rx)=r^af(x)$. We are also assuming (*) that if $\|x\|=\|y\|=1$ then $f(x)=f(y)$. We want to deduce that there is a constant $c$ such that for all $y$, $f(y)=c\|y\|^a$. 
As above, pick any $x$ with $\|x\|=1$, and define $c$ by $f(x)=c$. 
Since we are trying to prove something about all $y$, we let $y$ be an arbitrary vector. We know we can write $y=\|y\|v$ where $\|v\|=1$. Now let's calculate $f(y)$: 
$f(y)=f(\|y\|v)=\|y\|^af(v)$ by (#). Since $\|x\|=\|v\|=1$ and $f(x)=c$, (*) tells us that $f(v)=c$. So we get $f(y)=c\|y\|^a$, and we win. 
