Find the coefficient of `x^9` in the expansion of `(x^2-1/(3x))^9`.

Explanation

To find the coefficient of \( x^9 \) in the expansion of \( (x^2 - \frac{1}{3x})^9 \), we can follow these steps:

### Step 1: Identify the general term in the binomial expansion
The binomial expansion of \( (p + q)^n \) is given by the general term:
\[ T_{r+1} = \binom{n}{r} p^{n-r} q^r \]
In our case, \( p = x^2 \), \( q = -\frac{1}{3x} \), and \( n = 9 \). Therefore, the general term becomes:
\[ T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(-\frac{1}{3x}\right)^r \]

### Step 2: Simplify the general term
Now, we simplify the general term:
\[ T_{r+1} = \binom{9}{r} (x^{18 - 2r}) \left(-\frac{1}{3}\right)^r (x^{-r}) \]
Combining the powers of \( x \):
\[ T_{r+1} = \binom{9}{r} \left(-\frac{1}{3}\right)^r x^{18 - 3r} \]

### Step 3: Set the exponent of \( x \) equal to 9
We need to find the value of \( r \) such that the exponent of \( x \) is 9:
\[ 18 - 3r = 9 \]
Solving for \( r \):
\[ 18 - 9 = 3r \implies 9 = 3r \implies r = 3 \]

### Step 4: Substitute \( r \) back into the general term
Now, we substitute \( r = 3 \) back into the general term to find the coefficient:
\[ T_{4} = \binom{9}{3} \left(-\frac{1}{3}\right)^3 x^{18 - 9} \]
Calculating the coefficient:
\[ T_{4} = \binom{9}{3} \left(-\frac{1}{3}\right)^3 \]

### Step 5: Calculate \( \binom{9}{3} \) and \( \left(-\frac{1}{3}\right)^3 \)
Calculating \( \binom{9}{3} \):
\[ \binom{9}{3} = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \]
Calculating \( \left(-\frac{1}{3}\right)^3 \):
\[ \left(-\frac{1}{3}\right)^3 = -\frac{1}{27} \]

### Step 6: Combine the results to find the coefficient
Now, we multiply these results to find the coefficient of \( x^9 \):
\[ \text{Coefficient} = 84 \times \left(-\frac{1}{27}\right) = -\frac{84}{27} = -\frac{28}{9} \]

### Final Answer
The coefficient of \( x^9 \) in the expansion of \( (x^2 - \frac{1}{3x})^9 \) is:
\[ \boxed{-\frac{28}{9}} \]