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Is it possible to have a coordinate system $f:M\to R^n$ with $df_1\wedge...\wedge df_n(p)=0$?

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No, because then it wouldn't be a coordinate system. Every coordinate system induces a basis $\partial f_i$ for $T_pM$ (for each $p$ in the chart neighborhood), which induces a basis $df_i$ for $T_p^*M$, which induces the basis $df_1\wedge \cdots\wedge df_n$ for $\Lambda^n(T_pM)$. If $df_1\wedge\cdots\wedge df_n = 0$ anywhere, then it's not a basis for the one-dimensional vector space $\Lambda^n(T_pM)$.

From another perspective, $f$ is a diffeomorphism of the chart neighborhood onto $\mathbb{R}^n$, hence it induces an isomorphism between $\Lambda^n(T_pM)$ and $\Lambda^n(\mathbb{R}^n)$ for each $p$. The basis $df_1\wedge\cdots\wedge df_n$ is the pullback basis of the standard basis $e_1\wedge\cdots\wedge e_n$ of the $n$th exterior algebra of $\mathbb{R}^n$, and that's not zero, so neither can $df_1\wedge\cdots\wedge df_n$ be zero.