The solutions of the differential equation $y’=frac{x+y}{x-y}$, are given implicitly by the relation $$ln x = arctanfrac{y}{x}-frac{1}{2}ln(1+frac{y^2}{x^2})+c,enspace cinBbb{R}.$$

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 $ain A$, $bin B$ such that $f(a,b)=0$ where $A,B$ are open sets. Then for each $x_0in A$ there will be unique solution $y(x_0)in B$, which is differentiable and therefore continuous. Am I missing some key insights here?