Find the integral surface of linear PDE $(x^2-yz)p+(y^2-zx)q=z^2-xy$ which passes through the straight line $x=1, y=0$ Here $\;p=\frac{\partial z}{\partial x}\;$ and  $\;q=\frac{\partial z}{\partial y}$

Lagrange's equations are 
$
\begin{align}
\frac{dx}{x^2-yz} = \frac{dy}{y^2-zx} = \frac{dz}{z^2-xy}
\end{align}$

Let the general solution be $\phi (C1,C2) = 0 $

By Choosing multipliers $\; x,y,z\;$ and $\;1,1,1\;$ we get $\;C1\;$ as below.

\begin{align}
\frac{xdx+ydy+zdz}{x^3+y^3+z^3-3xyz} &amp; = \frac{dx+dy+dz}{x^2+y^2+z^2-xy-yz-zx} \\
\end{align}

\begin{align}
\frac{xdx+ydy+zdz}{(x+y+z)(x^2+y^2+z^2-xy-yz-zx)} &amp; = \frac{dx+dy+dz}{x^2+y^2+z^2-xy-yz-zx} \\
\end{align}

\begin{align}
\frac{xdx+ydy+zdz}{x+y+z} &amp; = dx+dy+dz \\
\end{align}

\begin{align}
xdx+ydy+zdz &amp; = (x+y+z)d(x+y+z) \\
\end{align}

\begin{align}
\frac{x^2}{2}+\frac{y^2}{2}+\frac{z^2}{2} &amp; = \frac{(x+y+z)^2}{2} + C \\
\end{align}

\begin{align}
x^2+y^2+z^2 - (x+y+z)^2 &amp; = C1
\end{align}

Now I am not able to find $C2$. The answer given in the textbook is $(x-y)(xy+yz+zx)+y-z = 0$

Thanks for any hint.
By Choosing multipliers $(y-z),(z-x),(x-y)$ :
$$\frac{dx}{x^2-yz} = \frac{dy}{y^2-zx} = \frac{dz}{z^2-xy} =
\frac{(y-z)dx+(z-x)dy+(x-y)dz}{(y-x)(x^2-yz)+(z-x)(y^2-xz)+(x-y)(z^2-xy)}=\frac{(y-z)dx+(z-x)dy+(x-y)dz}{0}$$
This implies 
$$(y-z)dx+(z-x)dy+(x-y)dz=0$$
I suppose that you can take it from here.
$$(x^2-yz)p+ (y^2-zx)q= z^2-yx$$
$$\frac{dx}{x^2-yz}= \frac{dy}{y^2-zx} = \frac{dz}{z^2-yx}$$
$$\frac{dx-dy}{x^2-y^2+zx-yz} = \frac{dy-dz}{y^2-z^2+xy-zx}$$
$$\frac{d(x-y)}{(x+y)(x-y)+z(x-y)} = \frac{d(y-z)}{(y+z)(y-z)+x(y-z)}$$
$$\frac{d(x-y)}{(x-y)(x+y+z)} = \frac{d(y-z)}{(y-z)(x+y+z)}$$
$$\frac{d(x-y)}{x-y} = \frac{d(y-z)}{y-z}$$
Now integrating both side we get:
$$\ln(x-y)= \ln(y-z) + c_1$$
$$\ln(x-y)-\ln(y-z)= c_1$$
Similarly if you compare
$\frac{dx-dz}{x^2-yz-z^2+xy} = \frac{dy-dz}{y^2-zx-z^2+xy}$,
you get:
$$\ln(x-z) = \ln(y-z)+ c_2$$
$$\ln(x-z)-\ln(y-z)= c_2$$