Why a holomorphic function has a regular value? Let $f : \mathbb{C} \rightarrow \mathbb{C}$ be a holomorphic function.
$y \in\ \mathbb{C}$ is said to be a regular value of $f$ if $\forall x \in f^{-1}(y)$ there exist an open neighborhood $U$ of $x$ and an open neighborhood $V$ of $y$ such that $f|U$ is a diffeomorphism onto $V$.
Could someone explain me why such a function $f$ has a regular value ?
 A: I think you probably mean so that $f |U$ is a diffeomorphism onto $V$. (Just stating that there are neighborhoods that are diffeomorphic is not really a condition about $f$, and moreover it is always true...) 
A reason such a regular value exists is because of local normal form for holomorphic functions: For $f$ a holomorphic function, and any point $x$ its domain, there is a neighborhood $U$ of $x$ and $W$ of $f(X)$, and biholomorphisms $\phi : D \to U$, $\psi : W \to D$ so that $\phi(0) = x$ and $\psi(f(x)) = 0$ so that $\psi f \phi : D \to D$ is $z^n$for some integer $n \geq 0$. Here $D$ is the unit disc.
So, $f$ is a local diffeomorphism at $x$ iff $n = 1$, which happens iff $f'(x)$ is nonzero. Now, the image of $\{ x : f'(x) = 0 \}$ under $f$ is a proper subset of $\mathbb{C}$, for example because $\{x : f'(x) = 0 \}$ is a countable subset of $\mathbb{C}$ (it must be discrete unless $f$ is constant, and discrete subset of $\mathbb{C}$ must be countable). Hence we can find some point $y$ so that for all $x \in f^{-1}(y)$, $f'(x) \not = 0$. Hence (again from the local normal form) there is a neighborhood of $x$ carried diffeomorphically by $f$ onto a neighborhood of $y$.
Alternatively, you can argue using Sards theorem. (But this theorem on local normal form has huge bang / buck ratio within complex analysis of one variable.)
