I have consulted other sources, but the relevant ones have used notation that I am not entirely comfortable with. I'm told that generalizing this procedure is straightforward, but I find myself getting confused/being unsure about what I am doing.
I want to get the general solution of the PDE $u_{x} + yu_{y} + zu_{z} = 0$
Attempt:
The characteristic curves should be subject to the relations $\frac{dx}{1} = \frac{dy}{y} = \frac{dz}{z}$. To figure out what the characteristic curves/surfaces are here, I believe I should consider two separate relations among these.
I have first considered $dx = \frac{dy}{y}$ to get that $\ln{y} = x + C$ for some constant C. Since the function $u$ should be constant on this characteristic curve, I solve for $C = \ln{y} - x$.
Next I consider the relation $\frac{dy}{y} = \frac{dz}{z}$ and similarly get that $C_{2} = \ln{(y-z)}$.
From this, I believe that I can deduce $u(x,y,z) = f(\ln{(y)}-x, \ln{(y-z)})$ is a general solution.
Do I have the right idea?
Note that your second relation is false. The equation $dy/y = dz/z$ implies that $C_2 = y/z$ and hence:
$$ u = f(C_1,C_2) = f(\log{y} -x , y/z),$$
since the characteristics method also tells us that the fractions are also equal to $ du/0$ and hence $u = \text{const}$.
You can check that this solution satisfies the PDE taking into account that:
$$ \frac{\partial f}{\partial x_i} = \frac{\partial f}{\partial C_1} \frac{\partial C_1}{\partial x_i} + \frac{\partial f}{\partial C_2} \frac{\partial C_2}{\partial x_i}, \quad x_i = \{x,y,z\}$$
Hope this helps.
Cheers!