When simplified and expressed with negative exponents, the expression $(x + y)^{ - 1}(x^{ - 1} + y^{ - 1})$ is equal to:
$\textbf{(A)}\ x^{ - 2} + 2x^{ - 1}y^{ - 1} + y^{ - 2} \qquad\textbf{(B)}\ x^{ - 2} + 2^{ - 1}x^{ - 1}y^{ - 1} + y^{ - 2} \qquad\textbf{(C)}\ x^{ - 1}y^{ - 1}$
$\textbf{(D)}\ x^{ - 2} + y^{ - 2} \qquad\textbf{(E)}\ \frac {1}{x^{ - 1}y^{ - 1}}$
Note that $(x + y)^{-1}(x^{-1} + y^{-1}) = \dfrac{1}{x + y}\cdot\left(\dfrac{1}{x} + \dfrac{1}{y}\right) = \dfrac{1}{x + y}\cdot\dfrac{x + y}{xy} = \dfrac{1}{xy} = x^{-1}y^{-1}$. The answer is $\textbf{(C)}$.