Can a partial differential equation have two different solutions? Consider:
$$x^2p+y^2q=(x+y)z$$ where $p=\frac{\partial z}{\partial x}$ and $q=\frac{\partial z}{\partial y}$. Thus by Lagrange's Method
$$\frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{(x+y)z}$$
$$\Rightarrow \frac{dx}{x^2}=\frac{dy}{y^2}
\Rightarrow \frac{1}{x}-\frac{1}{y}=k$$
$$\Rightarrow \frac{dx}{x^2}=\frac{dz}{(x+y)z}
\Rightarrow -\frac{1}{x}-\frac{\ln(z)}{(x+y)}=c$$
$$\Rightarrow \phi (\frac{1}{x}-\frac{1}{y},x+y+x\ln(z))=c$$
Or if we explore further 
$$\frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-xyz(x+y)}\Rightarrow \frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-x^2yz-xy^2z}\Rightarrow yzdx+xzdy-xydz=0 \Rightarrow xyz=k$$
So can we even write $f(\frac{1}{x}-\frac{1}{y},xyz)=k$. So which one is right?
Soham
 A: You can't integrate $\mathrm dz/z$ to $\log z$ and treat $x$ and $y$ in the denominator as constant. Likewise, Raymond's answer treats $y$ as constant in integrating $y/x^2$. In the method of characteristics, since we're finding curves, there's only one independent variable at a time, and the others have to be expressed in terms of it when we want to integrate.
Thus, Raymond's answer is correct up to
$$
\frac{(x+y)\,\mathrm dx}{x^2}=\frac{\mathrm dz}z\;,
$$
but then we have to express $y$ in terms of $x$,
$$
y=\left(\frac1x-k\right)^{-1}\;,
$$
to obtain
$$
\left(\frac1x+\frac1x\frac1{1-kx}\right)\mathrm dx=\frac1z\mathrm dz\;.
$$
Integrating this yields
$$
z=c\frac{x^2}{1-kx}\;,
$$
and then substituting
$$
k=\frac1x-\frac1y
$$
yields
$$
z=cxy\;,
$$
which is readily confirmed to solve the given equation.
[Edit:]
As doraemonpaul rightly pointed out, that's not the general solution, since $c$ can be chosen independently for each value of $k$, so it should be
$$
z=c(k)xy=c\left(\frac1x-\frac1y\right)xy\;.
$$
A: I think that your work using the characteristics method was right at the start including :

$$\frac{dx}{x^2}=\frac{dy}{y^2} \Rightarrow \frac{1}{x}-\frac{1}{y}=k$$

But the implication here was wrong :

$$\frac{dx}{x^2}=\frac{dz}{(x+y)z} \Rightarrow
 -\frac{1}{x}-\frac{\ln(z)}{(x+y)}=c$$

because the $x$ appears at the denominator of $dz$ as well as the numerator of $dx$ so that I think this should be :
$$\frac{dx}{x^2}=\frac{dz}{(x+y)z}
\implies \frac{(x+y)\,dx}{x^2}=\frac{dz}z$$
with Joriki's correction and using $\ y=\dfrac 1{\frac 1x-k}=\dfrac x{1-kx}$ we get : 
$$\left(\frac 1x+\frac 1x+\frac k{1-kx}\right)dx=\frac{dz}z\implies 2\ln(x)-\ln(1-kx)-\ln(z)=C_0$$
$$\implies \ln\left(\frac {x^2}{1-kx}\right)-\ln(z)=\ln(xy)-\ln(z)=C_0$$
(we replaced by $y$ again to remove the $k$ constant)
$$\implies \phi\left(\frac{1}{x}-\frac{1}{y},\ \frac {xy}z\right)=C$$
(you had another error here : forgetting the denominator of the second parameter)
After that you wrote :

EDITED:
  Or if we explore further 
  $$\frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-xyz(x+y)}\Rightarrow
 \frac{yzdx+xzdy-xydz}{x^2yz+xy^2z-x^2yz-xy^2z}\Rightarrow
 yzdx+xzdy-xydz=0 \Rightarrow xyz=k$$

If you divide $\ yzdx+xzdy-xydz=0$ by $xyz\ $ you get :
$\dfrac {dx}x +\dfrac {dy}y -\dfrac {dz}z = 0\implies\dfrac {xy}z=C_1$
So that the other way to write the solution doesn't differ from the first one :
$$\phi\left(\frac 1x -\frac 1y,\ \frac {xy}z\right)=C$$
Because of the arbitrary character of $\phi$ other equivalent parameters could have been obtained/chosen. Like replacing one of the parameters by the product $\dfrac{y-x}z$ or something more elaborate.
A: Both are not right.
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dx}{dt}=x^2$ , letting $x(1)=-1$ , we have $-\dfrac{1}{x}=t$
$\dfrac{dy}{dt}=y^2$ , we have $-\dfrac{1}{y}=t+y_0=-\dfrac{1}{x}+y_0$
$\dfrac{dz}{dt}=(x+y)z=\left(-\dfrac{1}{t}-\dfrac{1}{t+y_0}\right)z$ , we have $z(x,y)=\dfrac{f(y_0)}{t(t+y_0)}=xy~f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)$
