Rudin inverse function theorem interpretation 
Theorem: Suppose f is a $\mathcal{L}'$-mapping of an open set $E \subset R^n$ into $R^n$, $f'(a)$ is invertible for some a $\in$ E, and $b=f(a)$. Then there exist open sets U and V in $R^n$ such that $a \in U,b \in V$, f is one-to-one on U, and $f(U)=V$.

What does $\mathcal{L}'$-mapping mean? I found it's a first class of a differentiable function, but I don't get it meaning.
 A: This is definition 9.20 on page 219 (in the 3rd edition). Rudin uses $\mathscr C$ instead of your $\mathcal L$. 
The idea is the following. You start with a differentiable function $f: E\to \mathbb{R}^n$, with $E\subset \mathbb R^n$. Given $x\in E$, $f'(x)\in L(\mathbb{R}^n,\mathbb{R}^n)$ is a certain linear operator (see definition 9.11).
Now, there is a norm on $L(\mathbb{R}^n,\mathbb{R}^n)$, as explained at the start of the chapter, so that $L(\mathbb{R}^n,\mathbb{R}^n)$ is in particular a metric space. 
We can therefore regard $f'$ as a map $f':E\to L(\mathbb{R}^n,\mathbb{R}^n)$ between metric spaces, and ask whether this new function is continuous (in the ordinary sense of chapter 4). When this $f'$ happens to be continuous, we say that $f$ is of class $\mathscr{C}'$. 
Since this definition is fairly abstract, Rudin introduces a more practical criterion to test whether a given function is of class $\mathscr{C}'$, in terms of the continuity of the partial derivatives (Theorem 9.21). In elementary multivariate calculus books, the statement of this theorem is usually regarded as the definition of being a class $\mathscr{C}'$ function (also written $C^1$). 
