Suppose that $\alpha=(\alpha_1,\cdots,\alpha_n)$, all $\alpha_j$ are positive rational number. A polynomial $f$ satisfying
$$ f(\lambda^{\alpha_1}x_1,\cdots,\lambda^{\alpha_n}x_n)=\lambda f(x_1,\cdots,x_n) $$
for all $\lambda>0$, which is called weighted homogeneous polynomial.
Now we focus on $q:=f|_{\mathbb{S}_{\alpha}^{n-1}}$, here
$$ \mathbb{S}_{\alpha}^{n-1}=\left\{x\in\mathbb{R}^n:|x|_{\alpha}=\left(\sum_{j=1}^n x_j^{\frac{N}{\alpha_j}}\right)^{\frac{1}{N}}=1\right\} $$
and $N$ is a positive rational number such that $\frac{N}{\alpha_j}$ are all even integers.
What can we say about the points $x_0\in \mathbb{S}_{\alpha}^{n-1}$ satisfying
$$ q(x_0)=dq|_{x_0}=0. $$
Here $dq$ is the differential of $q$, can we recognize such points directly from the expression of $f$ instead of computing $dq$?
Let $E$ be the set of such points, and $$ F=\{x\in\mathbb{R}^n: \nabla f(x)=0\}. $$ Then $$ E=\left\{\frac{x}{|x|_{\alpha}}:x\in F\backslash\{0\}\right\}, $$ where $|\cdot|_{\alpha}$ is the distance in the definition of $\mathbb{S}^{n-1}_{\alpha}$.
The proof uses Euler's theorem of the homogeneous function.