If $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ smooth, $ g(x,y)= x^3 + y^3$ and $g \circ f \equiv 0$, then $\det Df \equiv 0$

Question

Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ be a smooth function and $g: \mathbb{R}^2 \rightarrow \mathbb{R}$ be defined by $(x,y) \mapsto x^3 + y^3$. Assume that $g \circ f$ is identically $0$. Then I want to show that $\det Df$ is identically $0$.

I want to check if my solution is correct. This is what I've done:

Since $g \circ f \equiv 0$, we have $D(g(f(x,y))) = 0$ for all $x,y$. Then $D(g(f)=Dg(f)Df = \nabla g (f) Df = \nabla (f_1^3 + f_2^3) Df=(3f_1^2 \nabla f_1 + 3f_2^2 \nabla f_2) Df=0$ which means that for any $x,y$ either $f_1(x,y)=f_2(x,y)=0$ or $Df=0$. This implies that $\det Df \equiv 0$.

Answer 1

Your chain rule is wrong. Note that $$Dg(x,y) = 3(x^2,y^2),$$ therefore $$Dg(f) = 3(f_1^2, f_2^2).$$ You have a product of matrices. This means that if the product is zero you cannot conclude that at least one of them is zero.