Question

Consider the expansion of (x-1/6x^2)^9. Find the constant term in this expansion.

Answer 1

the general term is ...
term(r+1) = C(9,r) x^r ((-1/6)x^2)^(9-r)
= C(9,r) (-1/6)^(9-r) x^r x^(18-2r)
= C(9,r) (-1/6)^(9-r) x^(18-r)

You will have a constant term when it contains no variable x, that is, when 18-r = 0
so r = 18
But that makes no sense since 0 ≤ r ≤ 9

confirmed by Wolfram, notice every term contains an x term
https://www.wolframalpha.com/input/?i=expand+%28x-1%2F6x%5E2%29%5E9

check your question, fix it, then follow my steps

Answer 2

or, maybe
term(r+1) = C(9,r) x^r (-1/(6x^2))^(9-r)
= C(9,r) (-1/6)^(9-r) x^r x^(2r-18)
= C(9,r) (-1/6)^(9-r) x^(3r-18)
r = 6
C(9,6) (-1/6)^3 = -7/18