By expansion, it's easy to see both are true iff $x+y+z=xy+yz+zx$.

$\frac{yz-x^2}{1-x}=\frac{zx-y^2}{1-y} \implies yz-x^2-y^2z+x^2y=zx-y^2-x^2z+xy^2$ $\implies zy-zx+y^2-x^2+x^2z-y^2z+x^2y-xy^2=0$ $\implies (y-x)(z+y+x-xz-yz-xy)=0$ Thus, either $x=y$ or $x+y+z=xy+yz+xz$. But $x=y$ is untrue by the problem statement, thus $x+y+z=xy+yz+xz$. And so $yz-x^2=(1-x)(xy+yz+xz) \iff yz-x^2=xy+yz+xz-x^2y-xyz-x^2z$ $\iff x^2+xy+xz-x^2y-xyz-x^2z=0 \iff x+y+z-xy-yz-xz=0$, true.

See http://www.artofproblemsolving.com/Forum/viewtopic.php?p=473500&sid=b34b580a8efc72be0c9829b17d00781c#p473500

$\frac{yz-x^2}{1-x}=\frac{zx-y^2}{1-y}=\frac{zx-y^2-yz+x^2}{x-y}=x+y+z.$ Bye.. Sayantan....