24)
If f( x/3 ) = x^2 + x + 1 then f(x) must have the form a x^2 + b x + c.
So let f(x) = a x^2 + b x + c which gives f( x/3 ) = a x^2 / 9 + b x / 3 + c.
Identifying coefficients gives a = 9, b = 3 and c = 1.
Solving f( 3z ) = 7 yields z1 = -1/3 and z2 = 2/9 which sums to -1/9.