Let $\Omega$ be a $C^2$ domain and assume that $0 \in \partial \Omega$ and that $e_n$ is orthogonal to the boundary of $\Omega$ at $0$. Then in a neighbourhood of $0$, we can put \begin{equation} x_n=\psi(x_1,x_2,\dots,x_{n-1}) \end{equation} for a $C^2$ function $\psi$. Is this sufficient to conclude that \begin{equation} \psi_i(0)=0 \quad \mbox{para} \quad i=1,2\cdots, n-1? \end{equation} Notice that in the case where $\Omega = \{x^2+(y-1)^2=1\}$ the answer is affirmative. In fact,near zero, $$y(x)= -\sqrt[2]{1-x^2}+1, y_x(x)=-\dfrac{1}{2}\dfrac{-2x}{\sqrt[2]{1-x^2}}\ \ \mbox{and} \ \ y_x(0)=0.$$
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Yes, this is true. Indeed, $C^1$ regularity is enough. Since the function $$F(x) = x_n-\psi(x_1,\dots,x_{n-1})$$ vanishes on the boundary of $ \Omega$, its gradient is orthogonal to the boundary. This implies that $\nabla F(0)$ is parallel to $e_n$. Since $$F(0) = \langle -\psi_1(0),\dots,-\psi_{n-1}(0),1\rangle$$ it follows that $\psi_i(0)=0$ for $i=1,\dots,n-1$.
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