I would like to know if my solution to the following exercise is correct
Let $C$ be the upper cone: $$C=\big\{(x,y,z)\in\mathbb R^3:x^2+y^2=z^2,\, z\geq 0\big\}$$ (a) Prove that $C$ is a smooth manifold.
(b) Prove that $C$ is not a smooth submanifold of $\mathbb R^3$.
For (a) we follow the Nigel Hitchin's definition 5 of smooth manifold. Since the map: $$\begin{array}{rcll} \varphi:&C&\longrightarrow & \mathbb R^2\\ &(x,y,z)&\longmapsto &(x,y) \end{array}$$ is a bijection, the pair $(C,\varphi)$ is a (global) chart of $C$. In fact, $\varphi$ is an homeomorphism so $C$ must be Hausdorff and has a countable basis of open sets. This implies that $\mathcal A=\{(C,\varphi)\}$ is an atlas of $C$ and $C$ is a smooth manifold.
For (b) we follow the definition 12 of submanifolds. Suppose that the natural inclusion $i:C\hookrightarrow \mathbb R^3$ is smooth and consider $\alpha:(-\xi,\xi)\to C$ a smooth map in $C$ such that $\alpha(0)=(0,0,0)$. By hipothesis, the composition $i\circ\alpha$ must be a smooth path in $\mathbb R^3$. However the $z$-component of this map is defined by: $$z:t\mapsto \sqrt{x(t)^2+y(t)^2}$$ where $x=x(t)$ and $y=y(t)$ are the $x$-component and $y$-component of $\alpha$ respectively. It is clear that $z=z(t)$ is not smooth at $t=0$. Thus $i\circ\alpha$ cannot be smooth on its domain. This implies that $C$ cannot be a smooth submanifold of $\mathbb R^3$.
Many thanks!