Prove that the cone $S = \{(x,y,z) \in \Bbb R^3 ~|~x^2+y^2-z^2=0 \}$ is not a smooth surface

Question

Prove that the cone $S = \{(x,y,z) \in \Bbb R^3 ~|~x^2+y^2-z^2=0 \}$ is not a smooth surface. The book I am reading - Vector calculus by Peter Baxandall gives the following hint : (Let $S \subseteq \Bbb R^m$ be a smooth surface. Then there exists an open subset $E \in \Bbb R^m$ such that $f$ is smooth on $E$ and $S$ is a level set of the function $f:E \subseteq \Bbb R^m \rightarrow \Bbb R$)

Attempt: Suppose that $S$ is a smooth surface. Then, there exists a smooth function $f: E \subseteq \Bbb R^3 \rightarrow \Bbb R$ such that $$S=\{(x,y,z) \in E~|~f(x,y,z)=0 \}$$

Clearly $(0,0,0) \in S$ as $f(0,0,0)=0$

Consider the three differentiable paths $\alpha.\beta,\gamma$ in $S$ given by :

$$\alpha(t)=\big(t,0,t\big),~~ \beta(t)=\big(0,t,t\big),~~\gamma(t)=\big(t,t,\sqrt 2 t\big), t\in [-1,1]$$

We can see that $\forall t \in \Bbb R : \alpha(t),\beta(t),\gamma(t) \subseteq S$ because $ f( \alpha(t))= f(\beta(t))= f(\gamma(t))=0$

Then $\alpha'(t)=\big(1,0,1\big),~~ \beta'(t)=\big(0,1,1\big),~~\gamma'(t)=\big(1,1,\sqrt 2 \big)$. We can see that these vectors are linearly independent.

This is the direction specified in the book. How do I use this info to obtain a contradiction ? Intuitively, we should now obtain a contradiction by proving that $f$ is not a smooth function. Any hints on obtaining a contradiction will be deeply appreciated

EDIT: Problem Statement $Q.11$ in the textbook with the hint :

Answer 1

Let $$ g(u) = \begin{cases} \exp\left(\frac{-1}{u^2}\right) & x \ge 0 \\ 0 & x < 0 \end{cases} $$ Then (after some work) you can show that $g$ is smooth, and $g^{(n)}(0) = 0$ for $n = 1, 2, 3, \ldots$.

Now let $$ h(x, y, z) = x^2 + y^2 - z^2 $$

and $$ f(x, y, z) = g(h(x,y,z)^2) $$ Then it's easy to see that

1. $f(x, y, z) = 0$ implies that $h(x, y, z) = 0$ and

2. $f$ is smooth, by the chain rule.

What this means is that the proposed method of proof cannot succeed, because I've exhibited a smooth function on 3-space whose zero-set is exactly the cone that you're supposed to show is not a manifold.

How can this be? It appears to be a contradiction!

But that's because either you (in transcribing) or the author (in writing) forgot a critical component of the theorem: a manifold $M$ is always the zero set of a smooth function $f$ with the extra property that the gradient of $f$ is nonzero at every point of $M$.

In my example, the opposite is extremely true: the gradient of $f$ is zero at every point of $M$. I could have been more cautious and constructed something whose gradient was zero only at the origin (which is the only "bad point"), but it would have taken more effort and been less clear, so I didn't.