smc

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 $\boxed{x^2-y^2}$