Differential Equation $y(x)'=(y(x)+x)/(y(x)-x)$ can someone give me some tips on how to solve this differential equation.
I looked at the Wolfram solution which substituted $y(x)=xv(x)$. I'd know how to solve from there, but I have know idea why they did it in the first place, well why the algorithm did it in the first place. When do you substitue $y(x)=xv(x)$? 
What are other ways on solving it?
 A: Approach 1
$$y = v x \rightarrow y' = v + x v'$$
Substitute into original ODE. Looks like you know how to do this.
Why does this work?
To solve the equation $\tag 1 \dfrac{dy}{dx} = \dfrac{f_1(x, y)}{f_2(x,y)},$
where $f_1(x,y)$ and $f_2(x,y)$ are homogeneous functions of the same degree in $x$ and $y$, we use the following approach. Let 
$$f_1(x,y) = x^n \phi_1\left(\dfrac{y}{x}\right), ~f_2(x,y) = x^n \phi_2\left(\dfrac{y}{x}\right).$$
From $(1)$, we have
$$\dfrac{dy}{dx} = \dfrac{f_1(x, y)}{f_2(x,y)} = \dfrac{x^n \phi_1\left(\dfrac{y}{x}\right)}{x^n \phi_2\left(\dfrac{y}{x}\right)} = \dfrac{\phi_1\left(\dfrac{y}{x}\right)}{\phi_2\left(\dfrac{y}{x}\right)}$$
We can now write this as
$$\dfrac{dy}{dx} = f\left(\dfrac{y}{x}\right)$$
Now, if we say $v = \dfrac{y}{x} \rightarrow y = vx$ and we substitute, we get a separable equation of the form:
$$\dfrac{dv}{f(v) - v} = \dfrac{dx}{x}.$$
Approach 2
Let $x + y = z$.
We have $x + y = z \rightarrow 1 + \dfrac{dy}{dx} = \dfrac{dz}{dx} \rightarrow \dfrac{dy}{dx} = \dfrac{dz}{dx} - 1$.
For the numerator, we substitute $z$ and for the denominator, we substitute $y-x = z-2x$. Do you see how to get that last one?
Now, the ODE is separable.
Approach 3
Let $y-x = z$.
Follow what was done in approach 2.
A: Let's assume a equation scaling $x \to \alpha x$ and $y \to \beta y$. Under such scaling the equation becomes
$$
y'
=
{\alpha \over \beta}\,
{y + \left(\alpha/\beta\right)x \over y - \left(\alpha/\beta\right)x}
$$
Then, we can see the equation doesn't change its form whenever $\alpha = \beta$. It means $y/x$ doesn't change its form either. Then, a change of variables
$\left(~y \to {\rm f}~\right)$ like $y/x = {\rm f}\left(x\right)$ should simplify the original equation as other people ( see @Amzoti ) already showed.
For this simple equation, this analysis can be too pretentious. However, it illustrates a general technique that can be quite useful in more complicated cases.
See, for example,
Applications of Lie Groups to Difference Equations ( Differential and Integral Equations and Their Applications ) 
by Vladimir Dorodnitsyn.
