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