Question 984822
The INVERSE relation, {{{(3y/sqrt(6y^2-12))=x}}}


{{{9y^2/(6y^2-12)=x^2}}}


{{{(3y^2)/(2y^2-4)=x^2}}}


{{{3y^2=2xy^2-4x^2}}}


{{{3y^2-2xy^2=-4x^2}}}


{{{2xy^2-3y^2=4x^2}}}


{{{(y^2)(2x-3)=4x^2}}}


{{{y^2=(4x^2)/(2x-3)}}}


{{{y=0+- sqrt(4x^2)/sqrt(2x-3)}}},  two parts.  Upper part and a lower part.


{{{y=0+- (2x)/sqrt(2x-3)}}}


{{{highlight(y=0+- 2x*sqrt(2x-3)/(2x-3))}}}, still this inverse relation is two parts.
Domain is  {{{x>=3/2}}}.  Each branch depends on which one of the branches of the original function is used.


{{{graph(300,300,-2,10,-3,9,2x*sqrt(2x-3)/(2x-3))}}}


{{{graph(300,300,-2,10,-9,3,-2x*sqrt(2x-3)/(2x-3))}}}