Getting x in terms of y I have this equation:
$$\dfrac{x}{y} = \dfrac{y-x}{x}$$
How would I separate $x$ and $y$ in $x^2+xy-y^2=0$ ?
 A: $$x^2 + xy - y^2 = 0$$
$$\frac{x^2}{y^2} + \frac{x}{y} - 1 = 0$$
So let $t = \frac{x}{y}$, giving $t^2 + t - 1 = 0$. Can you solve this equation for $t$, giving you a relation between $x$ and $y$ as required?
A: If
$\dfrac{x}{y} = \dfrac{y-x}{x}, \tag 1$
then, assuming $x \ne 0 \ne y$, as we must for the equation to make sense, we may set
$\alpha = \dfrac{x}{y}, \tag 2$
and writing (1) as
$\dfrac{x}{y} = \dfrac{y}{x} - 1, \tag 3$
we find
$\alpha = \alpha^{-1} - 1, \tag 4$
or
$\alpha^2 = 1 - \alpha, \tag 5$
or
$\alpha^2 + \alpha - 1 = 0; \tag 6$
then the quadratic formula yields
$\alpha = \dfrac{1}{2}(-1  \pm \sqrt 5); \tag 7$
since we know no more about $x$ and $y$, the best we can do is, from (1), write
$x = \alpha y, \tag 8$
i.e.,
$x =  (\dfrac{1}{2}(-1  \pm \sqrt 5))y; \tag 9$
curiously enough, since
$(\dfrac{1}{2}(-1  + \sqrt 5))(\dfrac{1}{2}(-1  - \sqrt 5)) = -1, \tag{10}$
we may also write
$y = -(\dfrac{1}{2}(-1  \pm \sqrt 5))x. \tag{11}$
A: Consider your equation, as a quadratic equation with respect to the variable $y$ :
$$-y^2 + xy + x^2$$
handling the variable $x$ as a parameter. 
Then, it would be : 
$$D=b^2-4ac=x^2+4x^2=5x^2$$
It's easy to check that $\forall x\in \mathbb R, D\geq 0$, so you can safely express $y$ with respect to $x$ via the solution of the quadratic equation, without worrying about complex numbers.
Then :
$$y_{1,2} = \frac{-b \pm \sqrt D}{2a} \Rightarrow y_{1,2} = \frac{-x+\sqrt{5x^2}}{-2}$$
A: $$\frac{x}{y}=\frac{y-x}{x}$$
$$x^2=y(y-x)$$
$$x^2=y^2-yx$$
$$x^2+yx=y^2$$
$$x^2+yx+\left(\frac{y}{2}\right)^2=y^2+\left(\frac{y}{2}\right)^2$$
$$\left(x+\frac{y}{2}\right)^2=\frac{5}{4}y^2$$
$$x+\frac{y}{2}=\pm\sqrt{5}\frac{y}{2}$$
$$x=\frac{y}{2}(-1\pm\sqrt{5})$$
