Suppose $V$ is a continuous function $R_3 \rightarrow R $ of three coordinates $x,y,z$ e.g. $V(x,y,z)=x^2+y^3+z^4$.
Is it true that when $$ V(x,y,z) =0 \hspace{1cm}(1)$$ also $$ xV_x + yV_y + zV_z =0 ? \hspace{1cm}(2)$$
($V_x=\partial V/\partial x$ etc.)
The reason I ask is because when $(x,y,z)$ are homogeneous point-coordinates in a plane, then $(V_x,V_y,V_z)$ are the homogeneous line-coordinates of the tangent to the curve (1). This follows from $dV=0$ because $ dV = V_x dx + V_y dy + V_z dz$.
Equation (2) states that the tangent is incident with the point on the curve (1) which must be true of course to be a tangent.
When the assumption that (2) follows from (1) is wrong, then the ides that $(V_x,V_y,V_z)$ is the tangent is also wrong?
So I am confused about how to understand these ideas.
I have one idea. The function V has to fulfill a certain condition to be a function $V=0$ for homogeneous coordinates of a curve in a plane. Because in view of $$ (px, px, pz) \equiv (x,y,z),$$ it must fulfill the condition that: $$ V(px, px, pz) = 0 \leftrightarrow V(x,y,z)=0 .\hspace{1cm} (3) $$ Otherwise the same point has multiple values and V is not well behaved for homogeneous coordinates. So maybe this is a geometrical proof for a special situation of Euler's homogeneous function theorem.
Therefore the V that I mentioned is not a curve in R_2.
By looking at Wikipedia, I now see that the function $f(p) =V(px, px, pz)$ in one particular point must fulfill $f'(p)=0$ to be valid which is just equation (2) ....
Theorem (Eulers hommogeneous functions theorem):
$V \in \mathbb{R}[X,Y,Z]$ is a homogeneous polynomial of degree $d$ if and only if $$ x V_x + y V_y + z V_z = d\cdot V. $$
Your example polynomial is not homogeneous, so it does not satisfy formula (2) by the theorem. We can also check explicitly that for the values $x=1,y=-1,z=0$, we get $V(x,y,z)=1-1=0$, but $$ xV_x+yV_y+zV_z = 2x^2+3y^3+4z^4 =2+3 \cdot (-1) = -1 \neq 0. $$