I can show that the sphere $x_1^2+x_2^2+\ldots+x_n^2=1$ is an $(n-1)$-dimensional manifold by considering the map $f(x_1,\ldots,x_n)=x_1^2+\ldots+x_n^2$, and noticing that $1$ is a regular value of $f$, i.e. $Df(p)$ is surjective for all $p\in f^{-1}(1)$.
What about the quadric $x_1^2+x_2^2+\ldots+x_{n-1}^2=x_n^2$? I want to use the same trick, so I consider the map $g(x_1,\ldots,x_n)=x_1^2+\ldots+x_{n-1}^2-x_n^2$. Now, if $p=(p_1,\ldots,p_n)$, $Dg(p)$ is the $1\times n$ matrix $$\begin{bmatrix}2p_1 & 2p_2 & \ldots & -2p_n \end{bmatrix}.$$
Unfortunately, $Dg(p)$ is not surjective when $p=(0,0,\ldots,0)$.
How can I show that the quadric is a ($(n-1)$-dimensional?) manifold?