For all non-zero numbers $x$ and $y$ such that $x = 1/y$, $\left(x-\frac{1}{x}\right)\left(y+\frac{1}{y}\right)$ equals
$\textbf{(A) }2x^2\qquad \textbf{(B) }2y^2\qquad \textbf{(C) }x^2+y^2\qquad \textbf{(D) }x^2-y^2\qquad  \textbf{(E) }y^2-x^2$
Using substitution, we can substitute y into the equation in the first parentheses. 
Therefore, we'll get \[\left(x-\frac{1}{\frac{1}{y}}\right)\left(y+\frac{1}{y}\right)\]
Because $x = \frac{1}{y}$, we can also see that $y = \frac{1}{x}$. Using substitution again, we can substitute x into the second equation getting
\[\left(x-\frac{1}{\frac{1}{y}}\right)\left(y+\frac{1}{\frac{1}{x}}\right)\]
Simplifying, we get $(x-y)(y+x)$. Multiplying, we get $\boxed{\textbf{(D)   }x^2-y^2}$
~awin