Note that $(x+y)^{2} = x^{2}+y^{2}+2xy = 3xy \Rightarrow x^{2}+y^{2} = xy$. Then,  $$\frac{x+y}{y}-\frac{x-y}{x}=\frac{x(x+y)-y(x-y)}{xy}=\frac{{{x}^{2}}+xy-xy+{{y}^{2}}}{xy}=\frac{{{x}^{2}}+{{y}^{2}}}{xy} = 1$$. 
Two non-zero real numbers, $x$ and $y$, satisfy $(x+y)^{2}=3xy$. Which of the following is a possible value of $$\frac{x+y}{y}-\frac{x-y}{x}$$? (Adapted From 2000 AMC 12 Problems, Question \#11) 
Options:
1. $-1$ 
2. $-\frac{1}{2}$ 
3. $\frac{1}{2}$ 
4. $1$ 
5. $$2$$ 