What is zweibein for a hypersurface $f(\pmb{x})=0$? For an N-dimensional hypersurface $f(\pmb{x})=0$ embedded in N+1 flat dimensional space in which the coordinates  $x^1,x^2,\dots,x^{N+1}$ are N+1 Cartesian ones, we can introduce $N$ local parameters $u^1,u^2,\dots,u^N$ such that $\pmb{x}=\pmb{x}(u^1,u^2,\dots,u^N)$. Then, a simple question is, what is zweibein for this hypersurface $f(\pmb{x})=0$? Please give some textbooks for references.
Question 1: The terminology zweibein may not be so commonly used in mathematics textbooks. In physics textbooks, the termiology dyad is also used instead.
Question 2: The zweibein may be referred to as the tetrad defined by $\pmb{e}^a_μ=∂\pmb{x}(u^1,u^2,\dots,u^N)/∂u^μ$
See: https://en.wikipedia.org/wiki/Tetrad_formalism
 A: Most good books on general relativity and supergravity have reviews on tangent spaces and Vielbeine/Vierbeine, as these are needed to express fermions on the tangent spaces to a curved space. Consider Weinberg's classic Gravity text Ch 12.5. 
Here, I'll give you a simplest example of Zweibeine on a beach-ball, $S^2$ which might serve you when you learn the general theory.
The constraint is, of course, $f=x^2+y^2+z^2-1=0$, so $df/2=0=zdz+ydy+xdx$, hence 
$$
dz=-(x dx +y dy)/z,
$$
hence 
$$
dz^2=(xdx+ydy)^2/(1-x^2-y^2)=(xdx+ydy)^2/w, \qquad w\equiv 1-x^2-y^2 .
$$
Consequently,
$$
ds^2=g_{\mu\nu}dx^\mu dx^\nu \\
=dx^2+dy^2+(x^2 dx ^2 +2xydx dy +y^2dy)/w,
$$
so that 
$$
g_{\mu\nu}= \delta_{\mu\nu} +x^\mu x^\nu /w,
$$
and its inverse 
$$
g^{\mu\nu}= \delta_{\mu\nu} -x^\mu x^\nu ,
$$
$$
\det g_{\mu\nu}=1/w .
$$
The Zweibeine are basically the "square roots" of the metric,
$$g_{\mu\nu}=\delta_{ab} V_\mu^a V_\nu^b , \\
\delta^{ab}=g^{\mu\nu} V_\mu^a V_\nu^b , 
$$
$$
V_\mu^a=\delta_{\mu a} -\frac{x^\mu x^a}{1-w}\left (1+\frac{1}{\sqrt {w}}\right ) 
$$
$$V^{\mu a}=\delta_{\mu a} -\frac{x^\mu x^a}{1-w}(1+{\sqrt {w}}).
$$
The latter convert commuting gradients $\partial_\mu$ to non-commuting gradients on the tangent space, $V_a^\mu \partial_\mu$.
The positive square root of w may be equivalently supplanted by its negative square root, if desired.
Check them. 
In matrix form,
$$
g_{\mu\nu}=   \begin{pmatrix}
      1+x^2/w&xy/w\\
      xy/w&1+y^2/w
    \end{pmatrix},
$$
$$
g^{\mu\nu}=   \begin{pmatrix}
      1-x^2&-xy\\
      -xy&1-y^2
    \end{pmatrix},
$$
and check the square root of the metric is 
$$
V_{\mu}^{a}=   \begin{pmatrix}
      1-x^2v&-xyv\\
      -xyv&1-y^2v
    \end{pmatrix},  \qquad v\equiv (1+1/\sqrt{w})/(1-w).
$$
