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 $x^2+xy-y^2=0$ ?

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
$y = -(\dfrac{1}{2}(-1  \pm \sqrt 5))x. \tag{11}$
