How to solve Cauchy problem? I'm new to this problem. Here is the question.
$$(y+xz)z_x+(x+yz)z_y=z^2-1$$
Find the integral surface which the curves it passes are $y=1$ and $z=x^2$
By Lagrange system i found $u$ and $v$. We have
$$\frac{dx}{y+xz}=\frac{dy}{x+yz}=\frac{dz}{z^2-1}$$
$$\frac{dx+dy}{(x+y)(z+1)}=\frac{dz}{z^2-1}$$
Assume $x+y=t$, $dx+dy=dt$
$$\frac{dt}{t}=\frac{dz}{z-1}$$
If we integrate both sides we conclude
$$u=\frac{x+y}{z-1}=c_{1}$$
$$\frac{dz}{z^{2}-1}=\frac{ydx-xdy}{x^{2}-y^{2}}=\frac{\frac{ydx-xdy}{y^{2}}}{(\frac{x}{y})^{2}}$$
Assume $\dfrac{x}{y}=k$. Then we conclude.
$$v=\log\left(\frac{z+1}{z-1}\right)+\frac{2y}{x}=c_{2}$$
Am I on the right path so far? what to do next?

I do not understand how you obtained the characteristic $v$. Here is my solution $$\frac{dx-dy}{(x-y)(z-1)}=\frac{dz}{z^2-1}\Rightarrow v=\frac{x-y}{z+1}=c_{2}$$
To find the integral surface which the curve it passes we should parametrize the curve, let $x=t$ then $z=t^2$ and $y=1$. Substituting these values into $c_1$ and $c_2$ we get
$$\frac{t+1}{t^2-1}=c_1, \quad\frac{t-1}{t^2+1}=c_2.  $$
Eleminating $t$ from the above equations
$$\Rightarrow t=\frac{c_1+1}{c_1}$$
We get $$\Phi(c_1,c_2)=\frac{\frac{c_1+1}{c_1}-1}{(\frac{c_1+1}{c_1})^2+1}-c_2=0.$$ Now if we plug $c_1=\frac{x+y}{z-1}$ and $c_2=\frac{x-y}{z+1}$ into $\Phi$ we can get the solution.
