I am trying to complete the following problem:
Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ be a smooth function such that $df \neq 0$ when $f = 0$. Let $\alpha$ be the 1-form defined by $$\alpha : = \frac{dx}{\partial f/\partial y}$$ on the submanifold $P : =\{ f = 0 \> , \> \partial f / \partial y \neq 0 \}$. Show that $\alpha$ extends to a smooth 1-form on the submanifold $N:=\{ f = 0 \}$.
My thought was to augment the denominator of the expression to something that is non-vanishing on all of $N$ but still restrict to $\alpha$ on $P$. In this spirit, I thought something along the lines of $$ \widetilde{\alpha} = \frac{dx}{\partial f/\partial x + \partial f/\partial y}. $$ This is certainly smooth on $N$ since $df \neq 0$ when $f = 0$ (i.e. on $N$) by hypothesis and $\widetilde{\alpha} = \alpha$ when $\partial f /\partial x = 0$ , but $\widetilde{\alpha} \neq \alpha$ when both $\partial f /\partial x , \partial f / \partial y \neq 0$.
I thought trying to bring in a partition of unity subordinate to the open sets $\{ \partial f / \partial x \neq 0\}$ and $\{ \partial f / \partial y \neq 0 \}$ might by worthwhile, but I have not made much progress. Any suggestions or hints about how to proceed?
Since $df \neq 0$ on $N = \{f=0\}$, we divide $N$ into
$$ N =N_x \cup N_y = (\{ f_x \neq 0\}\cap N) \cup (\{ f_y\neq 0\}\cap N).$$
The one form $\alpha$ is clearly defined on $\{ f_y\neq 0\}\cap N$. We need only to find another one form $\beta$ defined on $\{ f_x \neq 0\}\cap N$ so that $\beta = \alpha$ on the intersection.
By symmetry we would try
$$ \beta = -\frac{dy}{\partial f/ \partial x}.$$
(it turns out that $\frac{dy}{\partial f/ \partial x}$ is off by a sign, thus this choice) To see that $\beta = \alpha$ on the intersection, pick any point $p \in N_x\cap N_y$ and $X$ the tangent vector along $N$. Then $X = c (\partial f/\partial y, - \partial f /\partial x)$ for some constant $c$. Then
$$\alpha(X) = \beta (X) = c.$$
Thus $\alpha =\beta $ on the intersection.