1. We are asked to find the term independent of $x$ in the expansion of $\left(\frac{1}{2x} - \frac{x^2}{3}\right)^9$. 2. Consider the general term in the binomial expansion of $(a-b)^9$ given by: $$\binom{9}{k} a^{9-k} (-b)^k$$ where $a = \frac{1}{2x}$ and $b = \frac{x^2}{3}$. 3. The general term is: $$T_{k+1} = \binom{9}{k} \left(\frac{1}{2x}\right)^{9-k} \left(-\frac{x^2}{3}\right)^k = \binom{9}{k} \frac{(-1)^k}{2^{9-k} 3^k} x^{-(9-k)} x^{2k}$$ 4. Simplify the power of $x$: $$x^{-(9-k)} x^{2k} = x^{-9+k+2k} = x^{-9+3k}$$ 5. We want the term independent of $x$, so set the exponent of $x$ to zero: $$-9 + 3k = 0$$ $$3k = 9$$ $$k = 3$$ 6. Substitute $k=3$ into the general term coefficient: $$\binom{9}{3} \frac{(-1)^3}{2^{6} 3^{3}} = 84 \times \frac{-1}{64 \times 27} = 84 \times \frac{-1}{1728} = -\frac{84}{1728} = -\frac{7}{144}$$ 7. Therefore, the term independent of $x$ in the expansion is $-\frac{7}{144}$.