Let $p=x+y+z,$ $q=xy+yz+zx,$ and $r=xyz.$ Clearly $r\ne 0.$ We car rewrite the given equations as $(x-y)(x+y)=\frac{z-y}{yz},$ $(x-z)(x+z)=\frac{x-y}{xy},$ and $(y-z)(y+z)=\frac{x-z}{xz}.$ Because $x\ne y\ne z\ne x,$ by chasing around signs for the differences $x-y,x-z,y-z$ we see that a solution exists if one of the parameters is positive (say $x>0$) and the other two negative, or one is negative (say $x<0$)and the other two positive. Specifically, we need $x^2<|y|^2<a<|z|^2$ in the former case and $z^2<a<|x|^2<y^2$ in the latter case. The important thing is that in both cases $a>0,$ and $pr<0$ (and now we see why we could have $a^2=-pr$). Now, multiplying the expressions above we get $(x+y)(y+z)(z+x)=-\frac{1}{(xyz)^2},$ or $pq=\frac{r^3-1}{r^2}. \ (1)$ Adding the original three expressions we get $3a=(x^2+y^2+z^2)-\frac{xy+yz+zx}{xyz},$ and thus $\frac{2r+1}{r}\cdot q=p^2-3a,$ or, by using $(1),$ $3(apr)r^2=(pr)^3-(2r+1)(r^3-1).\ (2)$ Multiplying the original three expressions we get $a^3(xyz)=(x^2y-1)(y^2z-1)(z^2x-1)$ and after some manipulations we get $a^3r^2=(r^2+r+1)^2+apr(r+1).\ (3)$. [Note: we have $xz^2+x^2y+y^2z=a(x+y+z)+3,$ and from $(1)$ we get $x^2z+y^2x+z^2y=-2r-\tfrac{1}{r^2}-ap-3.$] Finally, by squaring and adding the original three expressions we get $3a^2=(x^2+y^2+z^2)^2-2(x^2y^2+y^2z^2+z^2x^2)+\frac{x^2y^2+y^2z^2+z^2x^2}{x^2y^2z^2}-2\left(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\right),$ or $3a^2=(p^2-2q)^2-2(q^2-2pr)+\frac{q^2-2pr}{r^2}-\frac{2}{r}\left(-\frac{p}{r^2}-ap^2-3p-q^2-pr\right).\ (4)$ [Note: In order to find the value of $x^3z+y^3x+z^3y$ we use the fact that $(x^2z+y^2x+z^2y)(x+y+z)=x^3z+y^3x+z^3y +(q^2-2pr)+pr.$] From $(1)$ and $(2)$ we get $p^2-2q=3a+\frac{r^3-1}{pr^3},$ and $q^2-2pr=-\frac{1}{p^2r^4}\left[6(apr)r^4+(r^3-1)(r+1)(3r^2-r+1)\right].$ Substituting into $(4)$ and making sure we keep the quantity $apr$ as our unknown, we get the following quadratic: $$3r^2(r+1)(apr)^2+(r^2+r+1)(6r^3+2r^2-r-1)(apr)+(r-1)(r^2+r+1)^2(3r^2+2r+1)=0.$$ The two solutions are $(apr)_{-}=-(r^2+r+1)$ and $(apr)_{+}=-(r^2+r+1)\frac{(r-1)(3r^2+2r+1)}{3r^2(r+1)}.$ It is easy to check that the $(apr)_{+}$ solution does not satisfy the system of equations $(2)$ and $(3)$ unless $r=1;$ however, in this case, $(apr)_{+}=0,$ which is not possible. On the other hand, the solution $(apr)_{-}=-(r^2+r+1)$ does. In the latter case we get from $(2)$ and $(3)$ that $a^3=r^2+r+1,$ $(pr)^3=-(r^2+r+1)^2,$ [and therefore $(qr)^3=-(r-1)^2(r^2+r+1)],$ and finally $a^2=-pr.$