Answer
$\large\frac{dx}{dy}=\frac{\sec x-x\sec^2 y}{\tan y-y\sec x\tan x}$
Work Step by Step
$y\sec x=x\tan y$ ____(1)
Differentiating (1) with respect to $y$ treating $x$ as dependent variable and $y$ as independent variable
$(y)^{'}\sec x+y(\sec x)^{'}=(x)^{'}\tan y+x(\tan y)^{'}$
$\sec x+y\sec x \tan x\large\frac{dx}{dy}$=$\large\frac{dx}{dy}$ $\tan y+x\sec^2 y$
$\bf Solving\;\; for$ $\large\frac{dx}{dy}$:-
$\sec x-x\sec^2 y=\large\frac{dx}{dy}$ $[\tan y-y\sec x\tan x]$
$\large\frac{dx}{dy}=\frac{\sec x-x\sec^2 y}{\tan y-y\sec x\tan x}$
Hence $\large\frac{dx}{dy}=\frac{\sec x-x\sec^2 y}{\tan y-y\sec x\tan x}$.
