North Macedonia

For $x=y=z=0$, we get $f(0,0)=(f(0,0){)}^2$ and because of $f(0,0)\ne 0$, we get that $f(0,0)=1$. For $x=y=0$ we get $f(0,0)=(f(0,z){)}^2$, from where $f(0,z)=\pm 1$, for every real number $z$. For $x=0$, we get $\pm 1=\pm f(y,z)$, so $f(y,z)=\pm 1$, for every real numbers $y$ and $z$.

Let $f(t,t)=-1$, for some real number $t$, then for $x=y=z=t$ we get that $f(-t,-t)=(-1{)}^2=1$, and this implies that for every real number $t$ either $f(t,t)=1$ or $f(-t,-t)=1$. Let $t$ an arbitrary real number and let $s$ be a real number such that $f(s,s)=1$ and $|s|=t$ (such a number exists, because if $f(t,t)=-1$, then for $s=-t$, $f(s,s)=1$). For $x=z=t$ and $y=-t$ we get $f(s,-sf(-s,s))=f(-s,s)$. If $f(-s,s)=-1$, then we get $f(s,s)=-1$, which is a contradiction, so from here $f(-s,s)=1$ and $f(s,-s)=f(-s,s)=1$. If $f(-s,-s)=-1$, for $x=s$ and $y=z=-s$ we get $$f(s,s)=f(s,-s)f(-s,-s)=-1,$$ which is not possible ($f(s,s)=1$). This implies that for every real number $t$ it holds $f(t,\pm t)=1$. If for some real numbers $u$ and $v$, it holds $f(u,v)=-1$. For $x=u$ and $y=z=v$, we get $f(-u,v)=-1$. For $x=z=v$ и $y=u$ we get $f(v,-u)=-1$. For $x=z=-u$ and $y=v$, we get $f(-u,-v)=-1$. For $x=-u$ and $y=z=-v$, we get $f(u,-v)=-1$. For $x=v$ and $y=z=-u$ we get $f(-v,-u)=-1$. For $x=z=-v$ and $y=-u$ we get $f(-v,u)=-1$ and for $x=z=v$ and $y=-u$, we get $f(v,u)=-1$. This implies that for every real numbers $u$ and $v$ for which $f(u,v)=-1$, the following relationships hold $$f(u,\pm v)=f(\pm u,v)=f(v,\pm u)=f(\pm v,u)=-1,$$ and if $f(u,v)=1$, the following relationships hold $$f(u,\pm v)=f(\pm u,v)=f(v,\pm u)=f(\pm v,u)=1$$ (if one is $-1$, than from the first case it follows that all are $-1$). From here we have $f(xf(x,z),yf(y,z))=f(x,y)$, for every real numbers $x,y$ and $z$, so $$f(x,y)=f(x,z)f(y,z).$$ If $f(a,b)=-1$, then for an arbitrary real number $z$ and $x=a,y=b$, we get $-1=f(a,z)f(b,z)$, so either $f(a,z)=1$ and $f(b,z)=-1$ or $f(a,z)=-1$ and $f(b,z)=1$. If for some real numbers $c$ and $d$, the equalities $f(a,c)=-1$ and $f(a,d)=1$ hold for $x=c,y=d$ and $z=a$ we get $$f(c,d)=f(c,a)f(d,a)=-1.$$ If $f(a,c)=f(a,d)=1$, for $x=c,y=d$ and $z=a$ we get $$f(c,d)=f(c,a)f(d,a)=1\text{ and if }f(a,c)=f(a,d)=-1,\text{ then for }x=c,y=d$$ and $z=a$ we get $f(c,d)=f(c,a)f(d,a)=1$, i.e if $f(a,c)$ and $f(a,d)$ have the same sign, then $f(c,d)=1$, and if they have the opposite sign, then $f(c,d)=-1$. Let for $M\subseteq {\mathbb{R}}^{+}\cup \{0\}$ hold $M=\{x\mid f(a,x)=1\}$ and for $N\subseteq {\mathbb{R}}^{+}\cup \{0\}$, hold $N=\{x\mid f(a,x)=-1\}$, then $M\cup N={\mathbb{R}}^{+}\cup \{0\}$ and $f(x,y)=1$ if and only if $|x|$ and $|y|$ belong to the same set, and $f(x,y)=-1$ if and only if $|x|$ and $|y|$ belong to different sets (if real numbers $a$ and $b$ such that $f(a,b)=-1$ don't exist, then the set $N$ is empty and $f(x,y)=1$, for all real numbers). Let $M$ be an arbitrary subset of ${\mathbb{R}}^{+}\cup \{0\}$ and $N={\mathbb{R}}^{+}\cup \{0\}\setminus M$. We define a function $f:{\mathbb{R}}^2\to \mathbb{R}$, for which $f(x,y)=1$ if and only if $|x|$ and $|y|$ belong to the same set, and $f(x,y)=-1$ if and only if $|x|$ and $|y|$ belong to different sets. For the function $f$, holds $f(xf(x,z),yf(y,z))=f(x,y)$ (the function does not depend on the sign before $x$ and before $y$). If $|x|$ and $|y|$ are in the same set, then $f(x,y)=1$ and $f(x,z)=f(y,z)$ (or $|z|$ is in the same set with $|x|$ and $|y|$, or in the other), so the product $f(x,z)f(y,z)=1=f(x,y)$. If $|x|$ and $|y|$ belong to different sets, then $f(x,y)=-1$, and $|z|$ is either in the same set with $|x|$ or with $|y|$, so $f(x,z)=-f(y,z)$, i.e. $f(x,z)f(y,z)=-1=f(x,y)$. This implies that, the functions defined in this way satisfy the given functional equation and from the discussion above, they are unique.