Answer
$\dfrac{(x^4-3x^2)}{(x^2-1)^2} \sec (\dfrac{x^3}{x^2-1}) \tan (\dfrac{x^3}{x^2-1}) $
Work Step by Step
Let us suppose that $f(x)=\sec (\dfrac{x^3}{x^2-1})$
We differentiate both sides with respect to $x$.
$f^{\prime}(x)=\dfrac{d}{dx} [\sec (\dfrac{x^3}{x-1})] \\=\sec (\dfrac{x^3}{x^2-1}) \tan (\dfrac{x^3}{x^2-1}) \times [\dfrac{(x^2-1) (3x^2)-x^3 \times 2x )}{(x^2-1)^2}]$
Simplify to obtain:
$f^{\prime}(x)=\sec (\dfrac{x^3}{x^2-1}) \tan (\dfrac{x^3}{x^2-1}) [\dfrac{(x^4-3x^2}{(x^2-1)^2}]=\dfrac{(x^4-3x^2)}{(x^2-1)^2} \sec (\dfrac{x^3}{x^2-1}) \tan (\dfrac{x^3}{x^2-1}) $
