Question:

**Step 1:** Write the general term of the binomial expansion. The general term $T_{r+1}$ in the expansion of $(A+B)^n$ is given by $T_{r+1} = \binom{n}{r} A^{n-r} B^r$. For the given expression $\left(\dfrac{2x^2}{y} + \dfrac{y}{3x}\right)^{12}$, we have $n=12$, $A=\dfrac{2x^2}{y}$, and $B=\dfrac{y}{3x}$. Thus, the general term is: $$T_{r+1} = \binom{12}{r} \left(\frac{2x^2}{y}\right)^{12-r} \left(\frac{y}{3x}\right)^r$$ **Step 2:** Simplify the general term. $$T_{r+1} = \binom{12}{r} (2)^{12-r} (x^2)^{12-r} (y)^{-(12-r)} (y)^r (3)^{-r} (x)^{-r}$$ $$T_{r+1} = \binom{12}{r} 2^{12-r} 3^{-r} x^{2(12-r)} x^{-r} y^{-(12-r)} y^r$$ $$T_{r+1} = \binom{12}{r} 2^{12-r} 3^{-r} x^{24-2r-r} y^{-12+r+r}$$ $$T_{r+1} = \binom{12}{r} 2^{12-r} 3^{-r} x^{24-3r} y^{2r-12}$$ **Step 3:** Determine the value of $r$ by matching the powers of $x$ and $y$ with $x^9 y^{-3}$. For the power of $x$: $$24 - 3r = 9$$ $$3r = 15$$ $$r = 5$$ For the power of $y$: $$2r - 12 = -3$$ $$2r = 9$$ $$r = 4.5$$ In a binomial expansion, the index $r$ must be a non-negative integer. Since the value of $r$ derived from the power of $x$ ($r=5$) is an integer, but the value of $r$ derived from the power of $y$ ($r=4.5$) is not an integer, and these values are inconsistent, the term $x^9 y^{-3}$ does not exist in the expansion. **Step 4:** State the coefficient. Since the term $x^9 y^{-3}$ does not exist in the expansion, its coefficient is $0$.