The answer is $f(x,a)\equiv c$ for some $c\in \{\pm 1,0\}$ Let $P(x,y,z,a,b,c)$ (variables in this order) denote assertion in the problem. Claim: if $f(x_0,a_0)=0$ for some $x_0, a_0$ then $f(x,a)\equiv 0$ Proof: suppose $x\ne x_0$ and $a\ne a_0$, then $P(x_0,x,2x-x_0,a_0,2a-a_0, a)$ yields $f(x,y)=0$. We can use the argument again to settle the case where one of $x=x_0, y=a_0$ holds. Otherwise, for all $x,a$, $f(x,a)\ne 0$ Claim: for all reals $x,y,a,b$, $f(x,a) f(y,b)=f(x,b) f(y,a)$ Proof: if $a=b$ or $x=y$ it's clear, so assume $a\ne b$ and $x\ne y$. Select $z$ arbitrarily and $c$ st $c(y-x)\ne ay+bz-az-bx$ or $by+az-bz-ax$ $P(x,y,z,a,b,c)$ $P(x,y,z,b,a,c)$ Yields $$f(x,a)f(y,b)=\frac{ f\left( \frac{x+y+z}{3}, \frac{a+b+c}{3} \right)} {f(z,c)} = f(x,b)f(y,a)$$ Therefore, $P(x,x-d,x+d, y-e, y, y+e)$ where $de\ne 0$ yields $f(x,y)=f(x,y-e)f(x-d,y)f(x+d,y+e)=f(x,y)f(x-d,y-e)f(x+d,y+e)$ Therefore $f(x-d,y-e)f(x+d,y+e)=1$. In other words, for all $y\ne x$ and $b\ne a$, $f(x,a)f(y,b)=1$ Now set $z\notin \{x,y\}, c\notin\{a,b\}$, we can see $f(y,b)f(z,c)=1$ and $f(z,c)f(x,a)=1$. This means $f(x,a)\in \{1,-1\}$ If $f(x,a)=1$ then $f(y,b)=1$ for all $y\ne x$ and $b\ne a$, and we similarly get $f(u,v)=1$ for all $u,v\in \mathbb{R}$. The case where $f(x,a)=-1$ follows from the same argument.