Step 1:

One asymptotes of the rectangular hyperbola is $x+2y-5=0$.The other asymptote is $\perp$ to it.The equation is of the form $2x-y+l=0.$

The combined equation is $(x+2y-5)(2x-y+l)=0$

Step 2:

The equation of the hyperbola differs only in the constant term.

Therefore its equation is of the form $(x+2y-5)(2x-y+l)=k$------(1)

Step 3:

The rectangular hyperbola passes through $(6,0)$ and $(-3,0)$.

Substituting these in equ(1) we get

$(6-5)(12+l)=k$

$\Rightarrow k-l=12$----(2)

$(-3-5)(-6+l)=k$

$\Rightarrow k+8l=48$----(3)

Step 4:

Solving eq(2) & eq(3) we get

$9l=36$

$l=4$

Substitute the value of $l$ in eq(2) we get

$k-4=12$

$k=12+4$

$k=16$

Step 5:

The equation of the rectangular hyperbola is $ (x+2y-8)(2x-y+4)=16$

On simplifying we get,

$2x^2+3xy-2y^2-6x+13y-36=0$

This is the required equation.