Step 1:
The centre of the hyperbola is $(2,1)$ and one of the asymptotes is $3x-y-5=0$.Since in an rectangular hyperbola,the asymptotes are $\perp$ to each other,the other asymptote is of the form $x+3y+k=0$.
It passes through the centre $(2,1)$
Therefore $2+3+k=0$
$\Rightarrow 5+k=0$
$\Rightarrow k=-5$
The second asymptote is $x+3y-5=0$
Step 2:
The combined equation of the asymptotes is $(3x-y-5)(x+3y-5)=0$
The equation of the rectangular hyperbola differs only in the constant term.Its equation is of the form $(3x-y-5)(x+3y-5)=k$
Step 3:
The point $(1,-1)$ lies on the rectangular hyperbola.
Therefore $(3+1-5)(1-3-5)=k$
$(-1)(-7)=k$
$\Rightarrow k=7$
The equation of the hyperbola is $(3x-y-5)(x+3y-5)=7$
On simplifying we get,
$3x^2+8xy-3y^2-20x-10y+18=0$