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 $(2,1)$
The second asymptote is $x+3y-5=0$
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 $(3x-y-5)(x+3y-5)=k$
The point $(1,-1)$ lies on the rectangular hyperbola.
The equation of the hyperbola is $(3x-y-5)(x+3y-5)=7$
On simplifying we get,