Solve $\frac{dy}{dx} - \frac{dx}{dy} = \frac{y}{x} - \frac{x}{y}$ Solve:
$$\frac{dy}{dx} - \frac{dx}{dy} = \frac{y}{x} - \frac{x}{y}$$
What I have done till now:
$$\left(\frac{dy}{dx}\right)^2  -1= \frac{dy}{dx}\left(\frac{y}{x} - \frac{x}{y}\right)$$
$$xy\left(\frac{dy}{dx}\right)^2  - xy= \frac{dy}{dx}\left({y^2} - {x^2}\right)$$
Now I am stuck.
EDIT: I have been through a course on ODEs (More like a course which added tools to our toolbox to solve ODEs) but it has been a long time since then. Everytime someone helps me with an answer it is an "Oh Yes! I forgot about that" moment. Is there any resource (Succinct PDF preferably) which can help me revise the tools used for solving ODEs? This looked good but I want more examples and methods.
 A: A hint: If $u$ and $v$ are real numbers $\ne0$ then
$$v-{1\over v}=u-{1\over u}$$
holds iff either $v=u$ or $v=-{\displaystyle{1\over u}}$.
A: $$xy\left(\frac{dy}{dx}\right)^2-\frac{dy}{dx}y^2+\frac{dy}{dx}x^2-xy=0$$
$$y\left(\frac{dy}{dx}\right)\left(x\left(\frac{dy}{dx}\right)-y\right)+x\left(x\left(\frac{dy}{dx}\right)-y\right)=0$$
$$\left(x\left(\frac{dy}{dx}\right)-y\right)\left(y\left(\frac{dy}{dx}\right)+x\right)=0 \Rightarrow$$
$$\frac{dy}{dx}=\frac{-x}{y} \Rightarrow \int y \,dy=-\int x\,dx  $$
$$\text{and}$$
$$\frac{dy}{dx}=\frac{y}{x} \Rightarrow \int \frac{dy}{y}=\int \frac{dx}{x}$$
A: With the hint from @christian-blatter:
$$
v-\frac{1}{v}=
u-\frac{1}{u}
$$
$$
\iff
u,v\neq0
\quad\text{and}\quad
v-u=\frac{1}{v}-\frac{1}{u}
=-\frac{v-u}{uv}
$$
$$
\iff
u=v\neq0
\quad\text{or}\quad
uv=-1
$$
(i.e. $v=\delta u^\delta\neq0$ for $\delta=\pm1$).
Now $\frac{dy}{dx}$ and $\frac{dx}{dy}$ are reciprocals, so
$$
\frac{dy}{dx} - \frac{dx}{dy} = \frac{y}{x} - \frac{x}{y}
$$
$$
\implies\qquad
\frac{dy}{dx}=\frac{y}{x}
\qquad\text{or}\qquad
\frac{dy}{dx}=-\frac{x}{y}
$$
$$
\implies\qquad
\int\frac{dy}{y}=\int\frac{dx}{x}
\qquad\text{or}\qquad
\int\;y\;dy=-\int\;x\;dx
$$
$$
\implies\qquad
\ln|y|=\ln|x|+c_1
\qquad\text{or}\qquad
\frac{y^2}{2}=\frac{x^2}{2}+c_2
$$
$$
\implies\qquad
y=\left(\pm e^{c_1}\right)x
\qquad\text{or}\qquad
y^2=x^2+2c_2
$$
$$
\implies\qquad
y=ax
\qquad\text{or}\qquad
y^2=x^2+b
$$
A: The differential equation reads
$$f\left(\frac{\partial y}{\partial x}\right)=f\left(\frac{y}{x}\right)$$
and 
$$\frac{\partial y}{\partial x}=\frac{y}{x}$$
has solution
$$y(x)=c\ x.$$
The kernel of the right hand side of the equation is a factor of $\pm 1$, which gets sucked up by $c$ already.
