Level curves, solving equation for $c$ I want to draw level curves of the following function
$$f:\mathbb R^2 \to \mathbb R,\quad f(x,y)=\begin{cases}\dfrac{xy}{x^2+y^2},&(x,y)\neq (0,0)\\0,&(x,y)=(0,0)\end{cases}$$
My first problem is to solve for y, if I try (and I am not sure if I can do that I) get:
$$y=\frac{x \pm \sqrt{x^2(1-4c^2)}}{2c}$$
Is there another way to solve for y?
If not, what about the $\pm$ sign?  
I can't have two different level curves for the same value of c; that just does not make any sense to me.
I would be so grateful for any tips or hints.

The level curves are the solutions of
$$x^2+2axy+y^2=0.$$
By completing the square,
$$(x+ay)^2-(a^2-1)y^2=0$$
or
$$(x+(a-\sqrt{a^2-1})y)(x+(a+\sqrt{a^2-1})y)=0$$ which is the equation of a pair of straight lines meeting at the origin.

In principle, you have at most one level curve through a point of the plane, but you can have as many curves as you want. F.i. consider $\sin x\sin y=\frac14$:
https://www.wolframalpha.com/input/?i=sin+x+sin+y%3D0.25
But at places where the function gradient cancels by forming a saddle point, you can have double points. F.i. consider $\sin x\sin y=0$:
https://www.wolframalpha.com/input/?i=sin+x+sin+y%3D0
