Consider the Lagrange equation $x^2 \frac{\partial z}{\partial x} + y^2 \frac{\partial z}{\partial y} = (x+y) z$. Then general solution of given equation is
$(1)$ $F(\frac{xy}{z}, \frac{x-y}{z}) = 0$, for any arbitrary differentiable function $F$.
$(2)$ $F(\frac{x-y}{z}, \frac{1}{x}-\frac{1}{y}) = 0$, for any arbitrary differentiable function $F$.
$(3)$ $z = F(\frac{1}{x}-\frac{1}{y})$, for any arbitrary differentiable function $F$.
$(4)$ $z = xy F(\frac{1}{x}-\frac{1}{y})$, for any arbitrary differentiable function $F$.
For this i just check option three by putting $z= \frac{1}{x}-\frac{1}{y}$. This doesn't satisfies the given equation . So i reject it . option 1,2,4 are made from each other . If i reject any one of them then i have to reject all. so in exam i tick them right . So according to me 1,2,4 is correct . Am i correct ?
Your argument is correct and well done.
There is no need for solving the PDE, but just for fun :
Characteristic system of ODEs : $\quad \frac{dx}{x^2}=\frac{dy}{y^2}=\frac{dz}{(x+y)z}$
First family of characteristic curves, from $\quad \frac{dx}{x^2}=\frac{dy}{y^2} \quad\to\quad \frac1x-\frac1y=c_1$
Second characteristic family of characteristic curves, from
$\frac{\frac{dx}{x}}{x}=\frac{\frac{dy}{y}}{y}=\frac{\frac{dx}{x}+\frac{dy}{y}}{x+y}=\frac{dz}{(x+y)z}\quad\to\quad \frac{dx}{x}+\frac{dy}{y}=\frac{dz}{z} \quad\to\quad \frac{z}{xy}=c_2$
General solution : $$\Phi\left(\frac1x-\frac1y \:,\: \frac{z}{xy}\right)=0$$ With $\Phi$ an arbitrary differentiable function of two variables. Or many other equivalent forms of implicit equations, such as $(1)$ and $(2)$ for example.
Or equivalently : $\quad\frac{z}{xy}=F\left(\frac1x-\frac1y \right)$ $$z=xy\,F\left(\frac1x-\frac1y \right)$$