# Proof that there is a point on a surface that is not in any tangent plane

Question give the surface $x^2$ $=$ $y(2+3x+z)$ and ask me to proof that there is a point on that surface that is not in any tangent planes of the surface

I put the graph in Wolfram Alpha and get that there is a double elliptical cone with one pointy-like point

I think that point is the answer with the same idea of 2D graph with a sudden twist having no tangent line

also, I did partial differentiation depending on each variable and get:

$f_x$ $=3y-2x$

$f_y$ $=2+3x+z$

$f_z$ $=y$

if I consider tangent plane formula of:

$f_x(x-a)+f_y(y-b)+f_z(z-c) = 0$

can I find the point by letting $f_x,f_y,f_z$ all equal to $0$ and conclude that the point that satisfies this condition be my point not being in any tangent plane

And can this be the proof?

Replace $z$ by $z-2$. The given equation can then be written as $$0=x^2-3xy-zy=\left(x-{3\over2}y\right)^2-\left({3\over2}y+{1\over3}z\right)^2+\left({1\over3}z\right)^2\ .$$ Letting $$u:=x-{3\over2}y,\quad v:={1\over3}z,\quad w:={3\over2}y+{1\over3}z$$ we obtain the equation $$w^2=u^2+v^2$$ in the $(u,v,w)$-coordinates, and this is indeed a double cone $C$. The origin $O$ is a singularity of $C$, and there is no tangent plane at $O$. But contrary to the statement in the title and in the body of the question, each and every tangent plane of $C$ passes through $O$.
(The original $(x,y,z)$-coordinates of $O$ are $(0,0,-2)$.