Let $f:R^n\to R^n$ be a smooth function and $g:R^n\to R$ $$g(x_1,\dots,g_n)=x_1^5+\dots+x_n^5$$ Assume that $g\circ f\equiv 0$. Prove that $\det Df\equiv 0$.
The only idea I have is to apply the chain rule $d(g\circ f)_p=dg_{f(p)}\cdot df_p$, where each term is the corresp. matrix of partial derivatives. Then since $g\circ f=0$, $d(g\circ f)_p=0$ for any $p$. But this doesn't imply that either of the two matrices on the RHS is zero. Is it still true that the determinant of one of them is zero? Why do we need the explicit form of $g$?