$\begin{align} & {{x}^{2}}-6xy+{{y}^{2}}=0\to {{x}^{2}}+{{y}^{2}}=6xy \\ & \frac{\left( x-y \right)\left( x-y \right)}{\left( x+y \right)\left( x+y \right)}=\frac{{{x}^{2}}+{{y}^{2}}-2xy}{{{x}^{2}}+{{y}^{2}}+2xy}=\frac{6xy-2xy}{6xy+2xy}=\frac{4}{8}=\frac{1}{2} \\ \end{align}$