Let $f(x)= x^3+ x$ and $g(x)= x^3-x$ for all $x \in \mathrm{R}.$ Find the derivative of composition function $g \circ f^{-1}$ at the point $x=2.$
Let $y= f(x)= x^3+ x$, then $y'= f' (x)= 3x^2+ 1$.
$f'(2)= 13 $ and the function itself at $x= 2$ will be $y (x=2)= 10$. By the inverse function theorem , $$ (f^{-1})' (y= 10) = \frac{ 1}{ f'(x=2)} = \frac{ 1} { 13} $$
And $(g \circ f^{-1})' = g'(f^{-1}(x)) (f^{-1}(x))'$
But how to evaluate $g'(f^{-1}(x))$, since directly finding the inverse of $f$ is not a better process here.
Taking from your last line of proof: $(g\circ f^{-1})'(2)=g'(f^{-1}(2))\cdot (f^{-1})'(2)= g'(1)\cdot \dfrac{1}{f'(f^{-1}(2))}=2\cdot \dfrac{1}{f'(1)}=2\cdot \dfrac{1}{4}= \dfrac{1}{2}$ .