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.
 A: 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
A: It might be better to express the function in polar coordinates; $\frac{xy}{x^2+y^2}$ becomes $\cos\theta\sin\theta=\frac12\sin2\theta$. This does not depend on $r$, so the level curve $\frac{xy}{x^2+y^2}=k$ is non-empty only for $k\in[-0.5,0.5]$ and consists of a pair of straight lines $\theta=\frac12\sin^{-1}2k$.

Your equation for $y$ in terms of $x$ and $c$ simplifies as
$$y=\frac{x \pm \sqrt{x^2(1-4c^2)}}{2c}=\frac{x\pm x\sqrt{1-4c^2}}{2c}=\frac{1\pm\sqrt{1-4c^2}}{2c}x$$
which yields two straight lines like the polar method above.
A: Another way to solve for $y$?  
Well, let's see . . . if we build upon what we have, that is, let
$\dfrac{xy}{x^2 + y^2} = c, \; \text{a constant}, \tag 1$
then find
$xy = cx^2 + cy^2, \tag 2$
from which we form the quadratic in $y$
$cy^2 - xy + cx^2 = 0; \tag 3$
we use the famous formula:
$y = \dfrac{x \pm \sqrt{x^2 - 4c^2x^2}}{2c} = \dfrac{x \pm \sqrt{x^2(1 - 4c^2)}}{2c} \tag 4$
as our OP Ang has done; taking another step,
$y = \dfrac{x \pm \vert x \vert \sqrt{(1 - 4c^2)}}{2c}; \tag 4$
now 
$\vert x \vert = \pm x, \tag 5$
depending on the sign of $x$; but since we already have a $\pm$ sign in the formula (4), it reduces to
$y = \dfrac{x \pm  x  \sqrt{(1 - 4c^2)}}{2c} = \dfrac{1 \pm \sqrt{1 - 4c^2}}{2c}x; \tag 5$
that is, the equation(s) for two straight lines through the origin.
