Consider a real polynomial $f(x,y,z)$ such that forall $t \in \mathbb{R} :$ $f(x,y,z)=0 \Rightarrow f(tx,ty,ty)=0$ If $f(x_0,y_0,z_0)=0$ then prove that forall $t \in \mathbb{R} : f(tx'+x_0,ty'+y_0,tz'+z_0)=t^n f(x'+x_0,y'+y_0,z'+z_0)$ , where $n=deg(f)$.
This tries to be a generalization of the same statement with the alternate hypothesis that $f(x,y,z)=0$ is a conical surface with singularity point $K(x_0,y_0,z_0)$
The claim is false. For a counterexample, consider $f(x, y, z) = x^2 + y^4$.
The polynomial $f$ satisfies the given condition: if $f(x, y, z) = 0$, then $x = 0$ and $y = 0$, therefore $f(tx, ty, tz) = f(0, 0, tz) = 0$.
Now, let $(x_0, y_0, z_0) = (0, 0, 0)$. Clearly $f(x_0, y_0, z_0) = 0$. But then, for $t = 2$ we have that $$f(tx' + x_0, ty' + y_0, tz' + z_0) = f(2x', 2y', 2z') = 4x'^2 + 16y'^4$$ $$t^n f(x' + x_0, y' + y_0, z' + z_0) = 16 f(x', y', z') = 16x'^2 + 16y'^4$$ which give different results e.g. for $(x', y', z') = (1, 0, 0)$.
