Why this special method introduce a nowhere vanishing form on $\mathbb{S^1}$? I'm studying "An Introduction to Manifolds, Loring W. Tu". In page 216 of it, Loring W. Tu introduces a method to find a nowhere vanishing form on $\mathbb{S}^1$ as follows:

To find a nowhere-vanishing 1-form on $\mathbb{S}^1$, we take the exterior derivative of both sides of the equation
  $$x^2 + y^2 = 1.$$
  Using the antiderivation property of $d$, we get
  $$2xdx+2ydy=0.(*)$$
  Of course, this equation is valid only at a point $(x,y) \in \mathbb{S}^1.$ Let $U_x =\{(x,y)∈\mathbb{S}^1|x \not=0\}$ and $U_y =\{(x,y)∈\mathbb{S}^1 |y\not=0\}.$ By $(*)$, on $U_x \bigcap U_y$:
  $$\frac{dy}{x} = -\frac{dx}{y}$$
  Define a 1-form $\omega$ on $\mathbb{S}^1$ by
  $$\omega =\begin{cases} 
       \frac{dy}{x}& on  \space U_x \\
      -\frac{dx}{y} &  on \space U_y\\
   \end{cases}
$$

I know this form is well defined and smooth, but I have two questions:
$1$-Why this method makes a "nowhere vanishing form"?
$2$-Suppose $f:\mathbb{R}^3\rightarrow \mathbb{R}$ be smooth and $f^{-1}(0)$ is a regular level set.How can I generalize this method for manifold $f^{-1}(0)$ and find a nowhere-zero 2 form?
 A: The $1$-form Tu defines is nowhere vanishing for the following reason. Consider $dx/y$ as a $1$-form on the subset of the circle where $y\ne 0$. It will vanish (as a $1$-form on $\Bbb S^1$) at the point $(a,b)\in\Bbb S^1$ only if the tangent space at $(a,b)$ is spanned by $\partial/\partial y$. But the equation $x\,dx+y\,dy=0$ on $\Bbb S^1$ tells you that this precisely when $y=0$, i.e., at points $(\pm 1,0)$, but we have restricted precisely to the complement of that set.
This method will generalize to hypersurfaces in arbitrary dimension, as long as you take the preimage of a regular value. For a surface $S$ given by $f(x,y,z)=0$, since $df\ne 0$ at every point of $S$, it follows that some partial derivative must be nonzero at every point of $S$. By analogy with what Tu gave for the circle, note that because $df|_S = 0$, we have 
$$\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz = 0$$
as a $1$-form on $S$. It follows that 
$$\frac{dy\wedge dz}{\partial f/\partial x} = \frac{dz\wedge dx}{\partial f/\partial y} = \frac{dx\wedge dy}{\partial f/\partial z}$$
whenever the expressions make sense. And at each point, (at least) one of them must be defined. You can check as I indicated for the circle case that  the $2$-form is nowhere-vanishing on $S$. (For example, $dy\wedge dz$ vanishes on $S$ only at a point where $\partial/\partial x$ is tangent to $S$, and this happens precisely when $\partial f/\partial x = 0$.)
