Let's consider $f : \mathbb{R}^n \rightarrow \mathbb{R}$ a differentiable function, and we claim that the restriction of $f$ to $S_1 = \{ (x_1,\dots, x_n) \in \mathbb{R}^n : x_1 + \dots + x_n = 1\}$ is constant. We want to show that $\exists x_0 \in \mathbb{R}^n \quad ||x_0|| < 1$ and $Df(x_0) = 0$.
So, as $f_{\big|S_1} : S_1 \rightarrow \mathbb{R}$ is constant, we have : $Df_{\big|S_1} : S_1 \rightarrow \mathcal{L}(\mathbb{R}^n, \mathbb{R})$ which verify for all $x \in S_1$, $Df_{\big|S_1}(x) = 0$. And as we know that $Df(x)$ is linear, we find that $Df(x)(h)=0$ for all $h \in \mathbb{R}^n$. But I want to show this property for an $x$ not in $S_1$ but an $x$ which verify : $||x|| < 1$. So I though Taylor's formula may help me, but I didn't find out how. All I know is that : $f(x+h)-f(x) = Df_x(h) + o(||h||)$ for all $x,h \in \mathbb{R}^n$.
Someone could help me ?
I don't think what you are trying to prove is true. Take $n=2$. If you consider the function $f(x_1,x_2)=x_1+x_2-1$. Then $f=0$ on $S_1$ but $\nabla f(x_1,x_2)=(1,1)\ne (0,0)$ everywhere.