Answer
$$ y' =-\frac{1}{2}\left( 1-\frac{1}{x^2}\right)\left( x+\frac{1}{x}\right)^{-3/2} .$$
Work Step by Step
Since $ y=\left( x+\frac{1}{x}\right)^{-1/2}$, by using the chain rule: $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$, the derivative $ y'$ is given by
$$ y'=-\frac{1}{2}\left( x+\frac{1}{x}\right)^{-3/2}\left( x+\frac{1}{x}\right)'
\\=-\frac{1}{2}\left( x+\frac{1}{x}\right)^{-3/2}(1-\frac{1}{x^2})\\ =-\frac{1}{2}\left( 1-\frac{1}{x^2}\right)\left( x+\frac{1}{x}\right)^{-3/2} .$$
