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

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 $z(x,y)=\dfrac{f(y_0)}{t(t+y_0)}=xy~f\left(\dfrac{1}{x}-\dfrac{1}{y}\right)$
