To find the term independent of x in the expansion of \(\left( x - \frac{1}{x^2} \right)^9\), we use the binomial theorem:

\(\left( x - \frac{1}{x^2} \right)^9 = \sum_{k=0}^{9} \binom{9}{k} x^{9-k} \left( -\frac{1}{x^2} \right)^k\)

The general term is:

\(T_k = \binom{9}{k} x^{9-k} \left( -\frac{1}{x^2} \right)^k = \binom{9}{k} (-1)^k x^{9-k-2k}\)

\(= \binom{9}{k} (-1)^k x^{9-3k}\)

We need the exponent of x to be zero for the term to be independent of x:

\(9 - 3k = 0\)

\(3k = 9\)

\(k = 3\)

Substitute \(k = 3\) into the general term:

\(T_3 = \binom{9}{3} (-1)^3 x^{9-3 \times 3}\)

\(= \binom{9}{3} (-1)^3 x^0\)

\(= \binom{9}{3} (-1)^3\)

\(= 84 \times (-1)\)

\(= -84\)