Problem Statement:-
If by eliminating $x$ between the equation $x^2+ax+b=0$ and $xy+l(x+y)+m=0$; a quadratic equation in $y$ is formed whose roots are same as those of the original quadratic equation in $x$, then prove that either $a=2l$, or $b=m$ or, $b+m=al$
Attempt at a solution:-
$$x^2+ax+b=0\tag{1}$$ $$xy+l(x+y)+m=0\tag{2}$$
As it is given that on eliminating $x$ from $(1)$ and $(2)$ we get a quadratic equation in $y$ which has the same roots as that of $x$ in equation $(1)$, so we can conclude that the solution for eq. $(2)$ is $x=y$.
Hence, eq. $(2)$ becomes $$x^2+2lx+m=0\tag{3}$$
As equations $(1)$ and $(2)$ have the same roots hence, we get $$\dfrac{1}{1}=\dfrac{a}{2l}=\dfrac{b}{m}$$
From this we get $$a=2l$$ and $$b=m$$.
I am not able to obtain the third relation, can you guide me in the right direction.
This is where you lost a set of solutions. It is true that the two equations have the same solutions, but $(1)$ is a quadratic and has two solutions. For $y$ to satisfy the same equation, it can be either $x$, or the other solution which is $(-a-x)$ by Vieta's formulas.
The case $y=x$ was addressed in the original question, giving $a=2l$ and $b=m$.
In the second case, substituting $y=-a-x$ in $(2)$ gives
$$x(-x-a) + l(x - x - a) + m = 0$$
$$-x^2 - a x - (al - m) = 0$$
Comparing coefficients to $(1)$ gives directly $al - m = b \iff b+m = al$.