Note that $x = 6-y \Rightarrow x+y = 6$. Then, $$ \frac{{{x}^{2}}}{x-y}+\frac{{{y}^{2}}}{y-x}=\frac{{{x}^{2}}-{{y}^{2}}}{x-y}=\frac{(x+y)(x-y)}{x-y}=x+y = 6$$. 
Two non-zero real numbers, $x$ and $y$, satisfy $x = 6-y$. Which of the following is a possible value of $$ \frac{{{x}^{2}}}{x-y}+\frac{{{y}^{2}}}{y-x}$$? (Adapted From 2000 AMC 12 Problems, Question \#11) 
(A) $-1$ 
(B) $-\frac{1}{2}$ 
(C) $\frac{1}{2}$ 
(D) $1$ 
(E) $$2$$ 