It's simply the product rule. You are differentiating $\displaystyle \frac{x+ u}{y}= (x+ u)\frac{1}{y}$ with respect to s. Presumably, x and y are functions of s but u, here, is independent of s so we can treat it as a constant. The derivative of $\displaystyle (x+ u)\frac{1}{y}$ is $\displaystyle \frac{x+ u}{dx}\frac{1}{y}+ (x+u)\frac{d\frac{1}{y}}{ds}= \frac{dx}{ds}\frac{1}{y}+ (x+ u)\left(-\frac{1}{y^2}\frac{dy}{ds}\right)$.
Sorry, I should have mentioned that all 3 functions, x, y and u are functions of s here.
I understand the product rule, but I don't have a clue where the (x+u)/y sprung from.
My understanding of this problem is that we're trying to find two functions
$\displaystyle \Phi(x,y,u) $ = constant
$\displaystyle \Psi(x,y,u) $ = constant
as the solution of the equation satisfies $\displaystyle F(\Phi,\Psi)$ = 0
Since $\displaystyle \frac{d}{ds}(\frac{x+u}{y})$ = 0,
(x+u)/y is a constant, and so there's $\displaystyle \Phi(x,y,u) $.
I just don't understand how we got there.
In the notes I have, there is alternative way of solving these quasilinear equations, which involves integrating the characteristic equations, then parametrising the initial line and setting s = 0. Does this sound familiar to anyone?