Finding when covector disappears

[Kreizhn]

Homework Statement

Find all the points $p \in \mathbb R^2$ such that $df_p =0$ where
$$df = \frac{ (y^2-x^2) dx - 2xy dy }{(x^2+y^2)^2}$$

The Attempt at a Solution

I figure the way this should be done is by solving the differential equation derived from
$$(y^2-x^2) dx - 2xy dy =0 .$$
It's either that or just find when the coefficients are identically zero. I can do either quite easily once I know for sure which I should be doing. Thoughts?

Edit: My only issue is that if I solve the differential equation, I will technically have a constant that I can't get rid of. Is there a canonical way of setting such a constant? Or is the solution the whole one-parameter family dictated by possible choices of the constant?

 

[Dick]

This a part of some larger problem, right? A covector is a linear functional on vectors, right? dx and dy are linearly independent covectors. The only way a*dx+b*dy can vanish is if a and b are both zero.

 

[Kreizhn]

Yeah sorry. For some reason I actually knew that but my brain died.

Indeed, for any covector field to be identically zero the components must be zero in any coordinate chart. Since $df_p(X) =0, \forall X \in T_pM$ we can, in particular, simply choose elements of the tangent basis, which forces the components to be zero.

I don't know why that slipped my mind. I appreciate the input though!