f(2x)=2(2x)-3=4x-3; but the inverse f^-1(2x) "plays" the function backwards, or "rewinds": the operation carried out last becomes first, so +3 is the inverse of -3, so that's f+3, then divide by 2, because division is the inverse of multiplication: (f+3)/2. Finally replace f with 2x: f^-1(2x)=(2x+3)/2.
f(f^-1(2x))=f(2x+3)/2)=[2(2x+3)/2]-3=2x+3-3=2x, so the product of the function and its inverse just returns the original argument. The answer checks out!