if $d_f \neq 0$ then f is a submersion I often see in my course of differential geometry that $d_f$ the differential of a smooth function $f$ is sumersive iff $d_f \neq 0$.
Is this a true statement, if yes, how could we prove that?
I didn't find any clues in the litterature.
Thank you.
 A: I guess you meant submersive not surjective.
I assume $f$ is a real-valued map from a manifold $M$, so that $\mathrm{d}f$ does not vanish if, and only, if $\mathrm{d}f_x$ is surjective for all $x$ in the manifold. Whence the result.
Remark. A linear form is surjective if, and only, if it is non-zero.
A: A linear map from a real vector space to $\mathbb R$ is surjective if and only if it is not the zero map.  I think what you're asking comes down to this. 
You didn't mention this, but I assume you mean $f$ to be a smooth function from a smooth manifold $M$ to the manifold $\mathbb R$.  To say that $f$ is a submersion is to say that the tangent space map $df_m:T_m(M) \rightarrow T_{f(m)}(\mathbb R)$ is surjective at every point $m \in M$.  
Each space $T_{f(m)}(\mathbb R) \cong \mathbb R$ is one dimensional, and your hypothesis is that each linear map $df_m$ is not the zero map, so there you go.
A: Suppose $f : M \to \Bbb{R}$ is a smooth real-valued function. Its differential $df_p$ (or $d_f$) is a linear map $df_p : T_pM \to T_{f(p)}\Bbb{R} \cong \Bbb{R}$ between tangent spaces, taking tangent vectors $v \in T_pM$ to vectors $df_p(v)$ (or a real number by identification $T_{f(p)}\Bbb{R} \cong \Bbb{R}$) at $f(p) \in \Bbb{R}$. A smooth map $f$ is called $\textbf{submersion}$ if $df_p$ is surjective for any $p \in M$.
Suppose $df_p : T_pM \to \Bbb{R}$ is the differential at $p$.

If $df_p$ is surjective, then it's obviously non-zero.
If $df_p $ is non-zero (means that not all $v \in T_pM$ mapped to $0$) then it must be surjective since if  $df_p(v_0)  = a \neq 0$, for some $v_0 \in T_pM$ , then any $c \in \Bbb{R}$ we have a vector $v :=\frac{c}{a} v_0$ such that $df_p(v) = c$.

