Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 24

Find $\displaystyle \frac{dx}{dy}$ of $y \sec x = x \tan y$ by Implicity Differentiation.
$\displaystyle \frac{d}{dy} (y \sec x) = \frac{d}{dy} (x \tan y) $


$
\begin{equation}
\begin{aligned}
(y) \frac{d}{dy} (\sec x) + (\sec x) \frac{d}{dy} (y) & = (x) \frac{d}{dy} (\tan y) + (\tan y) \frac{d}{dy} (x)\\
\\
(y) (\sec x \tan x) \frac{dx}{dy} + (\sec x)(1) &= (x) (\sec^2y) + \tan y \frac{dx}{dy}\\
\\
y \sec x \tan x \frac{dx}{dy} + \sec x &= x \sec^2 y + \tan y \frac{dx}{dy}\\
\\
y \sec x \tan x \frac{dx}{dy} - \tan y \frac{dx}{dy} &= x \sec^2y - \sec x\\
\\
\frac{dx}{dy} (y \sec x \tan x - \tan x) &= x \sec^2 y - \sec x\\
\\
\frac{\frac{dx}{dy} \cancel{(y \sec x \tan x - \tan x)}}{\cancel{y \sec x \tan x - \tan x}} &= \frac{ x \sec^2 y - \sec x}{y \sec x \tan x - \tan x}\\
\\
\frac{dx}{dy} &= \frac{ x \sec^2 y - \sec x}{y \sec x \tan x - \tan x}
\end{aligned}
\end{equation}
$