North Macedonia competition problems

For $x = y = z = 0$, we get $f(0,0) = (f(0,0))^2$ and because of $f(0,0) \neq 0$, 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.