An AP question asks:
The function f(x) = x^5 + 3x - 2 passes through the point (1,2). Let f^-1 denote the inverse of f. Then (f^-1)(2) equals?
The inverse of this function should be y^5 + 3y - 2. But, the book says that (f^-1)'(y) = 1 / f'(x).
Isn't there a discrepancy?
The inverse $y=g(x)$ of the function is (in principle) obtained by solving the equation $y^5+3y-2=x$. (Note that the function $x^5+3x-2$ has an inverse, since $x^5+3x-2$ is a strictly increasing function.) This is somewhat related to what you wrote, but different.
The equation $y^5+3y-2-x=0$ is "impossible" to solve explicitly for $y$ in terms of $x$. (I am oversimplifying). However, we can find the derivative of the inverse function of $f(x)$ at any point by using the formula you quoted.