Derivative function of inverse function

Let $g(x)$ be the derivative of the inverse function of $f(x) = \tan^2x$ for $0 \lt x \lt \frac{\pi}{2} .$ If the minimum value of $h(x)=\frac{-g'(x)}{\left(g(x)\right)^2}$ is $a,$ which is attained at $x=b,$ then what is $a^{2} + 60b?$