I have the following PDE
$$xu_x+u_y-(y+z)u_z=0$$
So I have to solve $$\frac{dx}{dt}=x,\frac{dy}{dt}=1,\frac{dz}{dt}=-(y+z),\frac{du}{dt}=0$$
I got $$x(t,s_1,s_2)=e^t\cdot f_1(s_1,s_2)$$ $$y(t,s_1,s_2)=t+f_2(s_1,s_2)$$
So for $\frac{dz}{dt}=-(y+z)$ I have to solve: $$z'+z=-t-f_2(s_1,s_2)$$
So
$z=z_h+z_p$
$$z'_h+z_h=0\rightarrow z_h=c\cdot e^{-t}$$
now $z_p=c(t,s_1,s_2)\cdot e^{-t}$
we derive and plug it back to the ODE $z'+z=-t-f_2(s_1,s_2)$ :
$$-c(t,s_1,s_2)\cdot e^{-t}+c'(t,s_1,s_2)\cdot e^{-t}+c(t,s_1,s_2)\cdot e^{-t}=-t-f_2(s_1,s2)$$
$$c'(t,s_1,s_2)\cdot e^{-t}=-t-f_2(s_1,s2)\Rightarrow c(t,s_1,s_2)=-te^{t}+e^{t}-f_2(s_1,s_2)e^{t}+f_3(s_1,s_2)$$
So $$z_p=(te^{t}+e^{t}-f_2(s_1,s_2)e^{t}+f_3(t,s_1,s_2))e^{-t}=t+1-f_2(s_1,s_2)+f_3(s_1,s_2)e^{-t}$$
So the solution is:
$$z=z_h+z_p=c\cdot e^{-t}+(te^{t}+e^{t}-f_2(s_1,s_2)e^{t}+f_3(t,s_1,s_2))e^{-t}=t+1-f_2(s_1,s_2)+f_3(s_1,s_2)e^{-t}$$
Now from what I know $z=z_h+z_p$ where $z_h$ is the solution of the homogeneous system and $z_p$ is the particular solution.
But in the end of the exercise it seems that the solution is only $z_p$
$$z=(t+1)+f_1(s_1,s_2)+e^t\cdot f_3(s_1,s_2)$$
What is correct?
You have correctly written $$\frac{dx}{dt}=x,\frac{dy}{dt}=1,\frac{dz}{dt}=-(y+z),\frac{du}{dt}=0$$ Or equivalently on the form of the Charpit-Lagrange system : https://en.wikipedia.org/wiki/Method_of_characteristics $$\frac{dx}{x}=\frac{dy}{1}=\frac{dz}{-(y+z)}=\frac{du}{0}=dt$$ A first characteristic equation comes from $\frac{dx}{x}=\frac{dy}{1}$ . Solving leads to : $$xe^{-y}=c_1$$ A second characteristic equation comes from $\frac{dy}{1}=\frac{dz}{-(y+z)}$ . Solving $\frac{dz}{dy}+z+y=0$ leads to : $$(y+z-1)e^y=c_2$$ A third characteristic equation comes from necessarily $du=0$ : $$u=c_3$$ The general solution of the PDE expressed on the form of implicit equation is : $$\Phi(c_1,c_2,c_3)=\Phi\big(xe^{-y}\:,\:(y+z-1)e^y\:,\:u\big)=0$$ where $\Phi$ is an arbitrary function of three variables.
Or equivalently $$\boxed{u(x,y,z)=F\big(xe^{-y}\:,\:(y+z-1)e^y\big)}$$ where $F$ is an arbitrary function of two variables.
It is easy to check that this equation satisfies the PDE with the calculus of $u_x$ , $u_y$ , $u_z$ and putting them into the PDE.
This shows that they are an infinity of solutions until some boundary condition be specified in order to determine the function $F$.
NOTE: Among the general solution found above, some less general solutions are $u=f(xe^{-y})$ and $u=g\big((y+z-1)e^y\big)$ where $f$ and $g$ are arbitrary functions.