When this formula working "smooth", Where from coming(with which logic)?Under what conditions working right?

Question

$$f(x_0)=y_0 \quad \Longleftrightarrow \quad f^{-1}(y_0)=x_0$$

$$\dfrac{d}{dx}(f^{-1}(y_0))=\dfrac{1}{f'(x_0)}$$

Answer 1

Suppose $f(x)$ is invertible and differentiable on some interval $[a,b]$. Write $y=f(x)$. Then $y'=f'(x)$. On the other hand, $f^{-1}(y)=x$, and differentiating both sides and using the chain rule gives:

$$\frac{d}{dx}[f^{-1}(y)]=\frac{df^{-1}(y)}{dy}\frac{dy}{dx}=\frac{df^{-1}(y)}{dy}f'(x)=1.$$

Rearranging the last equality gives the desired result.

Answer 2

Inverse Function Theorem: Let $f:\mathbb{R^n}\to\mathbb{R^n}$ be a $C^1$ function and $a\in\mathbb{R^n}$. If $A=f'(a)$ is invertible, that is, if $det(A)\neq 0$, then exists open sets $U,V\in\mathbb{R^n}$ such that $a\in U, f(a)\in V$ and $f:U \to V$ is a diffeomorfism. In that case: $$(f^{-1})'(f(x_0)) = (f'(x_0))^{-1}$$

In your case that work when $n=1$, if I understand what you mean.