Find $D_{x} y$. \[ y=(\sec x+\tan x)^{-3} \] $-3(\sec x+\tan x)^{-4}\left(\tan ^{2} x+\sec x \tan x\right)$ $-3(\sec x+\tan x)^{-4}$ $-3\left(\sec x \tan x+\sec ^{2} x\right)^{-4}$ $\frac{-3 \sec x}{(\sec x+\tan x)^{3}}$

Step 1 ：Given the function \(y=(\sec x+\tan x)^{-3}\), we are asked to find its derivative \(D_{x} y\).

Step 2 ：We can use the chain rule to find the derivative of this function. The chain rule states that the derivative of a composite function is the derivative of the outer function times the derivative of the inner function.

Step 3 ：In this case, the outer function is \(f(u) = u^{-3}\) and the inner function is \(g(x) = \sec x + \tan x\).

Step 4 ：We need to find the derivatives of these two functions and then multiply them together.

Step 5 ：The derivative of the function is \(-3(\sec x+\tan x)^{-4}(\tan ^{2} x+\sec x \tan x)\), which simplifies to \(-3(\sec x+\tan x)^{-4}\).

Step 6 ：Final Answer: The derivative of the function \(y=(\sec x+\tan x)^{-3}\) is \(\boxed{-3(\tan x+\sec x)^{-4}(\tan ^{2} x+\sec x \tan x)}\)