Equation $f(x,y) f(y,z) = f(x,z)$ How to solve the functional equation  $f(x,y) f(y,z) = f(x,z)$?
 A: Set $g(x)=f(x,0)$ and $h(z)=f(0,z)$. Then, we have $f(x,z)=f(x,0)f(0,z)=g(x)h(z)$ for all $x$ and $z$. Apply this to the original equation to obtain $g(x)h(y)g(y)h(z)=g(x)h(z)$.
There are three possibilities now:


*

*Either $g(x)=0$ for all $x$ (and thus $f(x,z)=0$ for all $x$ and $z$).

*Or $h(z)=0$ for all $z$ (and thus $f(x,z)=0$ for all $x$ and $z$ again).

*Or $g(x)\not=0$ and $h(z)\not=0$ for some $x$ and $z$. Then, we can cancel those two out to obtain $h(y)g(y)=1$ for all $y$. Thus, $h(y)=\frac{1}{g(y)}$ and we get $f(x,z)=\frac{g(x)}{g(y)}$. This can be easily checked to be a solution for any suitable $g(x)$.


It's easy to check there are two solutions: $f(x,y)=0$ for all $x,y$ and $f(x,y)=\frac{g(x)}{g(y)}$ with $g(x)$ being any function which is never equal to zero.
A: Suppose you know $f(x,z)= h_x(z)$ for some fixed $x$. Then you can work out
$$f(y,z) = f(x,z)/f(x,y)= h_x(z)/h_x(y)$$
That is, for all $y$ such that $h_x(y)\neq 0$ we find that $f$ is of the form $g(z)/g(y)$.
If $h_x(y)$ vanishes then $h_x(z)$ vanishes for all $z$. Then for any $a,b$ we have $$f(a,b)=f(a,x)f(x,b)=0$$
Hence the function is either identically zero, or nowhere zero and of the form $f(x,y)=g(y)/g(x)$ where $g$ is nowhere zero 
