The given function is \( \displaystyle y = \dfrac{1}{(1 + \tan^3 x)^2} \) and we need to find the derivative of this function with respect to \( x \).
Solution:
\( \begin{aligned} &\frac{dy}{dx} = \frac{d}{dx} \left[ \dfrac{1}{(1 + \tan^3 x)^2} \right] \\ &= \frac{d}{dx} (1 + \tan^3 x)^{-2} \\ &= -2(1 + \tan^3 x)^{-3} \cdot \frac{d}{dx}(1 + \tan^3 x) \\ &= -2(1 + \tan^3 x)^{-3} \cdot \left( \frac{d}{dx} 1 + \frac{d}{dx} \tan^3 x \right) \\ &= -2(1 + \tan^3 x)^{-3} \cdot \left( 0 + 3\tan^2 x \cdot \frac{d}{dx} \tan x \right) \\ &= -2(1 + \tan^3 x)^{-3} \cdot 3\tan^2 x \cdot \sec^2 x \\ &= -\frac{6 \tan^2 x \cdot \sec^2 x}{(1 + \tan^3 x)^3} \\ \end{aligned} \)Thus, the derivative of the function is:
\( \boxed{\frac{dy}{dx} = -\frac{6 \tan^2 x \cdot \sec^2 x}{(1 + \tan^3 x)^3}} \\ \)Please let me know in the comments if you find any errors in this solution.