I did see the solution to this question here but I still have doubts.
The subsidiary equation for the question is dx/(x+y) = dy/(z+x) = dz(x+y)
There are two methods to solving pde in the form of Pp + Qq = R (or at least the two we've been told to use.)
1) Method of grouping, where you take any two relations and solve them
2) Lagrange Multiplier, where you use the relation ldx + mdy + ndz/(lP + mQ + nR) and multiply terms such that the denominator is zero.
Using these two methods alone, I'm not able to arrive at the solution that is:
Does anyone know how I should proceed, can you please explain?
If the PDE is q(y+z)p + (z+x)q = x+y the "subsidiary equation" isn't dx/(x+y) = dy/(z+x) = dz(x+y) as you wrote but is : dx/(y+z) = dy/(z+x) = dz(x+y).
Search for a first characteristic equation : $$\frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dz}{x+y}=\frac{dx+dy+dz}{(y+z)+(z+x)+(x+y)}=\frac{dx-dy}{(y+z)-(z+x)}$$ $$\frac{dx+dy+dz}{2(x+y+z)}=\frac{dx-dy}{(y-x)}$$ $$\frac12\ln|x+y+z|=-\ln|x-y|+constant$$ Thus a first characteristic equation is : $$(x+y+z)(x-y)^2=c_1$$ Seach for a second characteristic equation : $$\begin{cases} \frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dx-dy}{(y+z)-(z+x)}=\frac{dx-dy}{(y-x)}\\ \frac{dy}{z+x}=\frac{dz}{x+y}=\frac{dy-dz}{(z+x)-(x+y}=\frac{dy-dz}{(z-y)} \end{cases}\quad\implies\quad \frac{dx-dy}{(x-y)}=\frac{dy-dz}{(y-z)}$$ $$\ln|x-y|=\ln|y-z|+constant$$ Thus a second characteristic equation is : $$\frac{x-y}{y-z}=c_2$$ General solution of the PDE on implicit form $f(c_1,c_2)=0$ : $$f\left((x+y+z)(x-y)^2\:,\: \frac{x-y}{y-z}\right)=0$$ $f$ is an arbitrary function of two variables (until bondary condition be specified).
In addition : ANSWER TO A QUESTION RAISED IN COMMENTS.
From the property of equal fractions $$\frac{A}{B}=\frac{C}{D}=\frac{E}{F}\quad\implies\quad \frac{A}{B}=\frac{C}{D}=\frac{E}{F}=\frac{c_1A+c_2C+c_3E}{c_1B+c_2D+c_3F}$$ With $A=dx\quad , \quad C=dy \quad , \quad E=dz \quad , \quad B=(y+z)\quad , \quad D=(z+x) \quad , \quad F=x+y \quad , \quad$ and with$\quad c_1=c_2=c_3=1$ $$\frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dz}{x+y}=\frac{dx+dy+dz}{(y+z)+(z+x)+(x+y)}$$ Also $$\frac{A}{B}=\frac{C}{D}\quad\implies\quad \frac{A}{B}=\frac{C}{D}=\frac{c_1A+c_2C}{c_1B+c_2D}$$ with $A=dx\quad , \quad C=dy \quad , B=(y+z)\quad , \quad D=(z+x) \quad , \quad c_1=1\quad$and$\quad c_2=-1$ $$\frac{dx}{y+z}=\frac{dy}{z+x}=\frac{dx-dy}{(y+z)-(z+x)}$$