$$\begin{cases} \frac{1}{x}z_x+\frac{1}{y}z_y=4\\ z(1,y)=y^2-1.\\ \end{cases}$$
So we started with:
$$\frac{dx}{dt}=\frac{1}{x}\rightarrow x^2(t,s)=2t+f_1(s)$$
$$\frac{dy}{dt}=\frac{1}{y}\rightarrow y^2(t,s)=2t+f_2(s)$$
$$\frac{dz}{dt}=4\rightarrow z(t,s)=4t+f_3(s).$$
From $$z(1,y)=y^2-1$$ we get $$x(0,s)=1,y(0,s)=s,z(0,s)=s^2-1.$$
How did we get $$x(0,s)=1,y(0,s)=s?$$
On
You could also write the initial data curve before analyzing the characteristic equations.
The initial data curve is $\Gamma: \langle 1,s,s^2-1 \rangle$. The characteristic equations are
$$\begin{cases} \dfrac{dx}{dt}=\dfrac{1}{x}, & x(0,s)=1 \\[1em] \dfrac{dy}{dt}=\dfrac{1}{y}, & y(0,s)=s \\[1em] \dfrac{dz}{dt}=4, & z(0,s)=s^2-1 \end{cases} $$
For the first equation,
$$\frac{dx}{dt}=\dfrac{1}{x} ~\Rightarrow~ x^2(t,s) = 2t+c_1(s) $$
Plugging in the initial condition of $ x(0,s)=1 ~\Rightarrow~ c_1(s)=1$. Hence,
$$x^2(t,s)=2t+1$$
For the second equation,
$$\frac{dy}{dt}=\dfrac{1}{y} ~\Rightarrow~ y^2(t,s) = 2t+c_2(s) $$
Plugging in the initial condition of $ y(0,s)=s ~\Rightarrow~ c_2(s)=s^2$. Hence,
$$y^2(t,s) = 2t+s^2$$
For the last equation,
$$\frac{dz}{dt}=4 ~\Rightarrow~ dz = 4dt ~\Rightarrow~ z(t,s)=4t+c_3(s)$$
Plugging in the initial condition of $ z(0,s)=s^2-1 ~\Rightarrow~ c_3(s)=s^2-1$. Hence,
$$ z(t,s)=4t+s^2-1$$
Combining all three equations together, we see that
$$x^2(t,s)=2t+1 \implies t=\frac{x^2-1}{2}$$ $$y^2(t,s) = 2t+s^2 \implies s=\sqrt{y^2-2t}=\sqrt{y^2-x^2+1}$$ $$z(t,s)=4t+s^2-1$$
So, the solution is
$$z=u(x,y)=4t+s^2-1=4\Big(\frac{x^2-1}{2}\Big)+\Big(y^2-x^2+1\Big)-1=x^2+y^2-2$$
On
When writing a characteristic system of this PDE $$\frac{dx}{dt} = \frac{1}{x},$$ $$\frac{dy}{dt} = \frac{1}{y},$$ $$\frac{dz}{dt} = 4,$$ then we can get constant functions on $t$ in the form $$I_{1}(x,y,z) = x^{2} - y^{2},$$ $$I_{2}(x,y,z) = 2x^{2} - z.$$ Now we are looking for an implicit function given by $$\phi(I_{1},I_{2}) = \phi(x^{2} - y^{2}, 2x^{2} - z) = 0.$$ After imposing the initial condition we have $$\phi(1 - y^{2}, 3 - y^{2}) = 0,$$ which implies that the possible form of the function $\phi(x,y,z)$ must be $$\phi(x,y,z) = x^{2} + y^{2} - z - 2.$$ Thus, solution of the PDE with the given IC is $$z(x,y) = x^{2} + y^{2} - 2.$$
This is follows by interpreting the initial condition $z(1,y)=y^2-1$ as $$z(x_0(s),y_0(s))=z_0(s).$$ Using $s=y$ as parameter of the underlying curve you get $x_0(s)=1$, $y_0(s)=s$, $z_0(s)=s^2-1$.
In the alternative formulation of the Lagrange equations they read as $$ x\,dx=y\,dy=\frac{dz}4 $$ so that along characteristic curves $y^2-x^2=c_1$ and $z-2x^2=c_2=\phi(c_1)=\phi(y^2-x^2)$ gives the general solution as $$ z(x,y)=2x^2+\phi(y^2-x^2). $$ Directly inserting the initial condition then gives $$ y^2-1=2+\phi(y^2-1)\implies \phi(u)=u-2\implies z(x,y)=x^2+y^2-2. $$