Show that $g\circ f\equiv 0$ implies $\det Df\equiv 0$, where $g(x_1,...,x_n)=x_1^5+...+x_n^5$

We know $g \circ f \equiv 0$. Therefore, by applying the chain rule we get $$Dg(f(x)) Df(x) \equiv 0.$$ Let us reason with the facts that $\operatorname{rank}(Dg(f(x))) \leq 1$ and $\operatorname{rank}(Df(x)) \leq n$.

We can't have $\operatorname{rank}(Dg(f(x))) = 1$ and $\operatorname{rank}(Df(x)) = n$ because this would mean the product is not zero.

If $\operatorname{rank}(Dg(f(x))) = 0$ then the whole $Dg(f(x))$ is zero. Designating the component functions of $f$ as $f_i$ we conclude that $f_i \equiv 0$ and $f \equiv 0$, leading to $\det(Df) \equiv 0$ because $f$ is the zero function.

If $\operatorname{rank}(Dg(f(x))) = 1$ then this means that $\operatorname{rank}(Df(x)) < n$. From this it follows that $\det(Df) \equiv 0$.