The solutions of the differential equation $y'=\frac{x+y}{x-y}$, are given implicitly by the relation $$\ln x = \arctan\frac{y}{x}-\frac{1}{2}\ln(1+\frac{y^2}{x^2})+c,\enspace c\in\Bbb{R}.$$
I'm considering the existence and uniqueness of arbitrary initial value problem $y(x_0) = y_0$. Let's say we write the equation as $f(x,y)=0$. The function will be continuously differentiable in $\Bbb{R^2}$. Then by the implicit mapping theorem, if we have points $a\in A$, $b\in B$ such that $f(a,b)=0$ where $A,B$ are open sets. Then for each $x_0\in A$ there will be unique solution $y(x_0)\in B$, which is differentiable and therefore continuous. Am I missing some key insights here?
Does this help:
One additional thing remains to do. Take $y=vx$, where $v$ is constant. The ODE gives $$v=\frac{1+v}{1-v} \implies v^2=-1, so this gives no real solution.
The general solution: Ley $y=v(x)x$, where $v(x)$ is non constant. then $y'=v+xv'$ then the ODE becomes $$v+xv'=\frac{1+v}{1-v}\implies xv'=\frac{1+v}{1-v}-v$$ $$\implies \int\frac{(1-v)dv}{1+v^2}=\int\frac{dx}{x}$$ $$\tan^{-1}v-\frac{1}{2}\ln (1+v^2)=\ln x +\ln C$$ $$\implies \tan^{-1}v=\ln Cx+\ln\sqrt{1+v^2}$$ $$\tan^{-1}(y/x)=D+\ln \sqrt{x^2+y^2}$$ which can be re written as $$\ln x=\tan^{-1}(y/x)-\frac{1}{2} \sqrt{1+y^2/x^2}+D$$