Answer
$-\displaystyle \frac{4}{27}$
Work Step by Step
$x=x(t)$ and we are given $x(1)=9$
By the chain rule,
$\left.\dfrac{dy}{dt}\right|_{t=1}=\left. \dfrac{dy}{dx}\dfrac{dx}{dt}\right|_{t=1}$... (evaluated at t=$1$)
$\left.\dfrac{dy}{dx}\right|_{t=1} = \left.\dfrac{dy}{dx}\right|_{x=9}\qquad$... because x($1$)=$9$
$\displaystyle \dfrac{dy}{dx}=\frac{d}{dx}[x^{1/2}+x^{-1/2}]=\frac{1}{2}x^{-1/2}-\frac{1}{2}x^{-3/2}$
$=\displaystyle \dfrac{1}{2\sqrt{x}}-\dfrac{1}{2x\sqrt{x}}$
$\displaystyle \left.\frac{dy}{dx}\right|_{x=9}=\frac{1}{2\cdot 3}-\frac{1}{2\cdot 9\cdot 3}=\frac{9-1}{54}=\frac{4}{27}$
$\displaystyle \left.\frac{dy}{dt}\right|_{t=1}=\left. \frac{dy}{dx}\frac{dx}{dt}\right|_{t=1}=\frac{4}{27}\cdot(-1)=-\frac{4}{27}$
