How do I solve a quasi-linear PDE using the Lagrange method?

[EAAL]

Homework Statement

Solve
(Z+e^x)Z_x + (Z+e^y)Z_y = Z^2 - e^{x+y}
Where Z = Z(x,y)

Homework Equations

Equations of the form

PZ_x + QZ_y = R
Where P = P(x,y,z) , Q=Q(x,y,z) , R=R(x,y,z)
Are solved with the Lagrange method.
It is possible to write this in the form:
\frac{dx}{P} = \frac{dy}{Q} = \frac{dz}{R}
Where from here we get two equations. We Solve these equations like ODE's, and the Final Solution would be F(c_1,c_2)=0 And C_1, C_2 are the integrating constants of the ODE's so C_1=C_1(x,y,z,c_2) and C_2=C_2(x,y,z,c_1)
It is possible to use the multiplier method, and try to find \alpha , \beta , \gamma (which can be functions)
To satisfy: \alpha P + \beta Q + \gamma R =0
Which would imply
(\alpha)dx + (\beta)dy + (\gamma)dz = 0
And to make this simple we look for \alpha = \alpha(x) , \beta=\beta(y) , \gamma=\gamma(z)

The Attempt at a Solution

(Z+e^x)Z_x + (Z+e^y)Z_y = Z^2 - e^{x+y}

\frac{dx}{z+e^x} = \frac{dy}{z+e^y} = \frac{dz}{z^2 - e^{x+y}}
But we can't get 2 solvable equations from here, because it would always involve the third variable, and we couldn't treat it like an ODE.
So we go to The Multiplier Mehod:
\alpha(z+e^x) + \beta(z+e^y) + \gamma(z^2 - e^{x+y}) =0
Which would imply:
\alpha + \beta + \gamma z = 0
and
\alpha e^x + \beta e^y - \gamma e^{x+y} = 0

But I haven't been able to find any \alpha , \beta , \gamma that do that...
I found \alpha = (z+e^y), \beta = (-z-e^x) , \gamma = 0
Fullfills, but then the equation
(z+e^y)dx - (z+e^x)dy = 0
Cannot be solved (we can't treat the z as a constant, because it is a function of x,y ...Thank you

 

[EAAL]

(*Bump*)