Suppose the two sheeted cone $S$ is a regular surface. As you note $p=(0,0,0)\in S$. Let $U\subset S$ be an open neighbourhood of $p$ in $S$ such that $U$ is the graph of a differentiable function of one of the forms
$$z=f(x,y),\qquad y=g(x,z),\qquad x=h(y,z).$$
By definition of the topologies on $S$ and $\Bbb{R}^3$ there exists an open ball $B_{\varepsilon}(p)\subset\Bbb{R}^3$, centered at $p$ and with radius $\varepsilon$, such that $V:=B_{\varepsilon}(p)\cap S$ and $V\subset U$. Then $V$ is also the graph of a function as above. Note that $V$ contains a point $(x,y,z)$ with $x\neq0$, $y\neq0$ and $z\neq0$. Then it also contains the points
$$(x,y,-z),\qquad (x,-y,z)\qquad\text{ and }\qquad (-x,y,z).$$
But this contradicts the fact that
$$z=f(x,y),\qquad y=g(x,z)\qquad\text{ or }\qquad x=h(y,z),$$
respectively. Hence the two sheeted cone is not a regular surface.
Hint.
Let us consider a cone instead of a truncated cone. The cone equation is given by
$$
C_{\lambda}(x,y,z) = x^2+y^2-\lambda^2z^2=0
$$
now rotating $\vec n = (0,0,1)$ around de axis $(0,1,0)$ by an angle $\theta$ we obtain $\vec n_{\theta} = (\sin\theta,0,\cos\theta)'$. After a tilt $\theta$ the liquid level can be described by the plane $P_{\theta}=\vec n_{\theta}\cdot(p-p_0)=0$ where $p = (x,y,z)'$ and $p_0 = (x_0,0,x_0/\lambda)'$. Now solving $\vec n_{\theta}\cdot(p-p_0)=0$ for $z$ we obtain $z = (\lambda+\tan\theta)x_0-x\tan\theta$ and the tilted cone volume is given by
$$
V_{\theta}=\int_{\Omega(x,y,\theta)}\left((\lambda+\tan\theta)x_0-x\tan\theta-\frac{1}{\lambda}\sqrt{x^2+y^2}\right)d\Omega
$$
Here
$$\Omega(x,y,\theta) = C_{\lambda}(x,y,z)\cap P_{\theta}=\cos ^2\theta \left(\lambda ^4 x_0^2-x^2-y^2\right)-\lambda ^2 (x-x_0) \sin \theta(2 \lambda x_0 \cos\theta+(x_0-x)\sin\theta ) \ge 0$$
Finally we can obtain the tilt angle by solving for $\theta$
$$
V_0 = \frac 13\frac{\pi}{\lambda} x_0^3=V_{\theta}
$$
Best Answer
Basically by the same reason why $\sqrt{1+x^2}$ is differentiable, whereas $\sqrt{x^2}(=\lvert x\rvert)$ is not: the surface $x^2=y^2+z^2$ is the union of the surfaces $x=\pm\sqrt{y^2+z^2}$ and you have a problem concerning differentiability when $(x,y,z)=(0,0,0)$. But you have no such problem with the surfaces $x=\pm\sqrt{1+x^2+y^2}$; each one of thm is well-behaved at any point, as far as differentiability is concerned.