Follow the method in http://en.wikipedia.org/wiki/Characteristic_equations#Example:
$(x^2+3y^2+3u^2)u_x-2xyuu_y+2xu=0$
$2xyuu_y-(x^2+3y^2+3u^2)u_x=2xu$
$2yu_y-\left(\dfrac{x}{u}+\dfrac{3y^2}{xu}+\dfrac{3u}{x}\right)u_x=2$
$\dfrac{du}{dt}=2$ , letting $u(0)=0$ , we have $u=2t$
$\dfrac{dy}{dt}=2y$ , letting $y(0)=y_0$ , we have $y=y_0e^{2t}=y_0e^u$
$\dfrac{dx}{dt}=-\left(\dfrac{x}{u}+\dfrac{3y^2}{xu}+\dfrac{3u}{x}\right)=-\dfrac{x}{2t}-\dfrac{3y_0^2e^{4t}}{2xt}-\dfrac{6t}{x}$
$\dfrac{dx}{dt}+\dfrac{x}{2t}=-\left(\dfrac{3y_0^2e^{4t}}{2t}+6t\right)\dfrac{1}{x}$
Let $w=x^2$ ,
Then $\dfrac{dw}{dt}=2x\dfrac{dx}{dt}$
$\therefore\dfrac{1}{2x}\dfrac{dw}{dt}+\dfrac{x}{2t}=-\left(\dfrac{3y_0^2e^{4t}}{2t}+6t\right)\dfrac{1}{x}$
$\dfrac{dw}{dt}+\dfrac{x^2}{t}=-\dfrac{3y_0^2e^{4t}}{t}-12t$
$\dfrac{dw}{dt}+\dfrac{w}{t}=-\dfrac{3y_0^2e^{4t}}{t}-12t$
I.F. $=e^{\int\frac{1}{t}dt}=e^{\ln t}=t$
$\therefore\dfrac{d}{dt}(tw)=-3y_0^2e^{4t}-12t^2$
$tw=\int(-3y_0^2e^{4t}-12t^2)~dt$
$tx^2=-\dfrac{3y_0^2e^{4t}}{4}-4t^3+f(y_0)$
$x^2=-\dfrac{3y_0^2e^{4t}}{4t}-4t^2+\dfrac{f(y_0)}{t}$
$x=\pm\sqrt{-\dfrac{3y_0^2e^{4t}}{4t}-4t^2+\dfrac{f(y_0)}{t}}$
$x=\pm\sqrt{-\dfrac{3y^2}{2u}-u^2+\dfrac{2f(ye^{-u})}{u}}$
$$xu_x+(1+y)u_y=x(1+y)+xu$$
Charpit-Lagrange equations :
$$\frac{dx}{x}=\frac{dy}{1+y}=\frac{du}{x(1+y)+xu}=ds$$
A first characteristic equation comes from solving $\frac{dx}{x}=\frac{dy}{1+y}$ :
$$\frac{1+y}{x}=c_1$$
A second characteristic equation comes from solving $\frac{dx}{x}=\frac{du}{x(1+y)+xu}=\frac{du}{x(c_1x)+xu}$
This is a first order linear ODE : $\frac{du}{dx}-u=c_1x$
$$ue^{-x}+c_1(x+1)e^{-x}=c_2$$
$c_1$ and $c_2$ are arbitrary related which leads to the general solution on the form of implicit equation :
$$ue^{-x}+\frac{1+y}{x}(x+1)e^{-x}=F\left(\frac{1+y}{x}\right)$$
$F$ is an arbitrary function (to be determined according to boundary conditions).
$$u(x,y)=-\frac{1+y}{x}(x+1)+e^xF\left(\frac{1+y}{x}\right)$$
FIRST CASE of boundary condition :
$u(x,6x-1)=\phi(x)=-\frac{1+(6x-1)}{x}(x+1)+e^xF\left(\frac{1+(6x-1)}{x}\right)$
$\phi(x)=-6(x+1)+e^xF\left(6\right)$
This implies that the function $\phi(x)$ must have a particular form :
$$\phi(x)=-6(x+1)+e^xC\quad\text{where}\quad C=\text{constant}$$
In general the functions $\phi(x)$ have not this very particular form and as a consequence the function $F$ cannot be determined. The problem has no solution fitting to the boundary condition. If by luck the function $\phi(x)$ has the above particular form, the solution is $\quad u(x,y)=-\frac{1+y}{x}(x+1)+e^xC$
SECONDCASE of boundary condition :
$u(-1,y)=\psi(y)=-\frac{1+y}{-1}(-1+1)+e^{-1}F\left(\frac{1+y}{-1}\right)=e^{-1}F(-1-y)$
Let $X=-1-y\quad;\quad y=-X-1$
$\psi(-X-1)=e^{-1}F(X)$
The function $F(X)$ is determined :
$$F(X)=e\:\psi(-X-1)$$
Now we can put it into the general solution where $X=\frac{1+y}{x}$ :
$$u(x,y)=-\frac{1+y}{x}(x+1)+e^x e\:\psi\left(-\frac{1+y}{x}-1\right)$$
In this case, the problem has a well determined solution fitting to the boundary condition :
$$u(x,y)=-\frac{1+y}{x}(x+1)+e^{x+1}\psi\left(-\frac{1+y+x}{x}\right)$$
Best Answer
Ok, I figured this out. It is clear that $y/z=\mathrm{const}=\psi$ will work. For $\phi$ we can substitute $z=y/c$. We end up with an exact equation:
$$2xy dx + (x^2+3y^2+\dfrac{3}{c^2}y^2) dy = \dfrac{d}{dy}(\phi(x(y),y)) = 0$$
with $$\phi=x^2y+y^3+\dfrac{yy^2}{c^2} = x^y+y^3+z^2y.$$
Clearly $\phi$ is a constant since its derivative is 0.
Thus: $F\left(\dfrac{u}{y},x^2y+y^3+u^2y\right)=0$ is the general solution for an arbitrary $F$ with continuous first derivatives. Using the implicit function theorem this is the same as:
$$u=yf(x^2y+y^3+u^2y).$$