jmc

We have $x+y+z=0,$ and squaring this equation gives $$(x^2+y^2+z^2)+2(xy+yz+zx)=0.$$Thus, $x^2+y^2+z^2=-2(xy+yz+zx).$ Since $x,y,z$ are distinct, it is not possible that $x^2+y^2+z^2=0,$ so we have $$\frac{xy+yz+zx}{x^2+y^2+z^2}=\boxed{-\frac{1}{2}}.$$