Answer
$5~ft/sec$
Work Step by Step
Here, $\dfrac{dy}{dt}=(-3e^{-x/3})(\dfrac{dx}{dt})$
As we are given that $\dfrac{dy}{dt}=\dfrac{-1}{4}\sqrt {9-y}~ft/sec$
Then, we have $\dfrac{dx}{dt}=\dfrac{e^{x/3}}{12}\sqrt {9-y} $
Now, at $(9,\dfrac{9}{e^3})$
$\dfrac{dx}{dt}=\dfrac{e^{9/3}}{12}\sqrt {9-(\dfrac{9}{e^3})} \approx~5 ft/sec$
