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

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.
