Answer
$\dfrac{dy}{dx}=-\dfrac{y^2}{{\sec{\dfrac{1}{y}}\tan{\dfrac{1}{y}}}}.$
Work Step by Step
$\dfrac{d}{dx}(x)=\dfrac{d}{dx}(\sec{\dfrac{1}{y}})\rightarrow$
Using the Chain Rule with $u=y^{-1}\rightarrow\dfrac{du}{dx}=-\dfrac{1}{y^2}\rightarrow$
$\dfrac{d}{dx}(\sec{\dfrac{1}{y}})=\dfrac{dy}{dx}(-\dfrac{\sec{\dfrac{1}{y}}\tan{\dfrac{1}{y}}}{y^2})\rightarrow$
$1=\dfrac{dy}{dx}(-\dfrac{\sec{\dfrac{1}{y}}\tan{\dfrac{1}{y}}}{y^2})\rightarrow$
$\dfrac{dy}{dx}=-\dfrac{y^2}{{\sec{\dfrac{1}{y}}\tan{\dfrac{1}{y}}}}.$
