Implicit derivative switching $y$ and $x$ Regard $y$ as the independent variable and $x$ as the dependent variable and use implicit differentiation to find $dx/dy$  if
$y \sec x = 4x \tan y$
I got $(\sec x-4x\sec^2 y)/4\tan y-y\sec \tan x$ but it was wrong.
 A: You almost got it.  
$\begin{align}
y\sec x & = 4x \tan y 
\\[1ex] 
y' \sec x + y \tan x \sec x & = 4 \tan y + 4 x y' \sec ^2 y & \text{implicit derivation using the chain rule}
\\[1ex]
y' (\sec x - 4 x \sec ^2 y) & = 4 \tan y - y \tan x \sec x & \text{associate like elements}
\\[1ex]
\frac{\mathrm d y}{\mathrm d x} & = \frac{4 \tan y - y \tan x \sec x}{\sec x - 4 x \sec ^2 y} & \text{cross division} 
\\[1ex]
\therefore \frac{\mathrm d x}{\mathrm d y} & = \frac{\sec x - 4 x \sec ^2 y}{4 \tan y - y \tan x \sec x} & \text{inverting} 
\end{align}$
Note though: You wrote:   $(\sec x−4 x \sec^2 y)/4\tan y−y\sec\tan x$
Not only is that   $\dfrac{\sec x - 4 x \sec ^2 y}{4 \tan y} - y \sec \tan x$
But also You have a typo:   $\;\sec \tan x \;$ is not the same as $\;\sec x \tan x\;$
A: differencing $y\sec x = 4x \tan y$ can be simplified to read $$y = 4x\cos x \tan y $$ differencing the latter gives you $$dy = 4x\cos x \sec^2 y \, dy + 4dx \cos x \tan y -4x \sin x \, dx \tan y \tag 1$$ 
that gives $$\frac{dx}{dy} = \frac{1-4x\cos x \sec^2 y}{4(\cos x \tan y - x \sin x \tan y)} \text{ and } (1)$$
