If a=0 multiple three equations and get xyz(xyz-1)=0 suppose xyz=0 and then x=0 then yz=0 then y=0 and setting in 3rd equayion get z=0. If xyz-1=0 then instead of yz,zx,xy put 1/x ,1/y and 1/z respectively solve and get x=y=z=±1.

If $z=0$ then $x=y=0$ If $z\neq 0$ thus \[x=ay+yz\quad y-az=z ay+yz\quad z-a =ay+yz\Leftrightarrow x=ay+yz\quad y=\frac {az}{1-az-z^2}\quad z-( ay+yz)=y (ay+yz)\]From $(2)$ and $(3) $ we get $z-a^2+az\cdot \frac {az}{1-az-z^2}=a+z\left (\frac {az}{1-az-z^2}\right)^2$ Restating the equation $z^5+a^2+2az^4+2a^3-2z^3+a^4-a^3-a^2-2az^2+1-a^3z=0$ We will prove that the equation has $5$ solutions. It's easy to see the equation has $2$ solutions. Which is $0$ and $1-a $ $(z) (z-1+a+z^3+a^2+a+1)( z^2+a^3-1)(z-a^2-a-1)=0$ Next, $f (z)=z^3+a^2+a+1+z^2+a^3-1+z-a^2-a-1$ By Rolle's theorem \[f (-\infty)<0, f(-2a)>0, f (0)<0, f(+\infty)>0\].