Let the subset of $\mathbb{R}^3$ be $S = \{(x,y,z) \in \mathbb{R}^3 : (1+\lvert x\rvert)^2-y^2-z^2=0\}$ and $T = S \cap \{(x,y,z) \in \mathbb{R}^3 : x \gt 0\}$
I want to show that T is a regular surface and S is not. Before proceeding to my proof, I have a some questions. In general, if S is a smooth surface, does it mean S is regular? Furthermore, would it be sufficient to show that the normal vector of a surface is never zero, to prove a surface is regular?
Now proving T is regular. Since $x \gt 0 \iff \lvert x\rvert = x$, then consider the maps $\sigma_{1,2}:\mathbb{R}^2 \to S \cap T$ such that $\sigma_1(x,y) = (x,y,\sqrt{(1+x)^2-y^2}) $ and $\sigma_2(x,y) = (x,y,-\sqrt{(1+x)^2-y^2})$. Then:
For smoothness, since the composition of smooth functions is smooth then both $\sigma_1(x,y)$ and $\sigma_2(x,y)$ are smooth.
For homeomorphism, we show bijection of $\sigma_i$ and continuity of $\sigma_i^{-1}$ and $\sigma_i$:
Injectivity: $(x_1,y_1,\pm \sqrt{(1+x_1)^2-y_1^2})=(x_2,y_2,\pm \sqrt{(1+x_2)^2-y_2^2}) \implies (x_1,y_1)=(x_2,y_2)$ so injective and hence bijective on its image which is $\mathbb{R}^2$
Continuity: all components in the range of $\sigma_i$, namely $(x,y,\pm \sqrt{(1+x)^2-y^2})$ are continuous thus $\sigma_i$ continuous
Now consider $\sigma_i^{-1}(x,y,\pm \sqrt{(1+x)^2-y^2}) =(x,y)$ for $i=1,2$ which exists and is simply a projection onto the $-xy$ plane with each component in the range, namely $(x,y)$ is continuous. Therefore $\sigma_i$ homeomorphisms from $\mathbb{R}^2$ to $T$
Finally, we check if the normal vector is ever $0$. We have $\frac{∂\sigma_i}{\partial x} = (1,0,\pm \frac{x +1}{\sqrt{(1+x)^2-y^2}})$ and $\frac{∂\sigma_i}{\partial y} = (0,1, \mp \frac{y}{\sqrt{(1+x)^2-y^2}})$ for $i=1,2$ resp.:
$\mathit{\overrightarrow N}= \frac{∂\sigma_i}{\partial x} \wedge \frac{∂\sigma_i}{\partial y} = (\pm \frac{x+1}{\sqrt{(1+x)^2-y^2}}, \mp \frac{y}{\sqrt{(1+x)^2-y^2}}, -1 ) \neq \mathbf{\overrightarrow 0}$
So the two partials are linearly independent elements of $\mathbb{R}^3$ and by the previous proofs, $T$ must be a regular surface.
How can I show that S is not regular? Would I encounter problems with the absolute value when trying to compute derivatives to show smoothness?