33. (1/9 axy²) · (-81a³x²y).

$$\left(\frac{1}{9}axy^2\right) \cdot (-81a^3x^2y) = \frac{1}{9} \cdot (-81) \cdot (a \cdot a^3) \cdot (x \cdot x^2) \cdot (y^2 \cdot y) = -\frac{81}{9} a^4x^3y^3 = -9a^4x^3y^3$$

Ответ: -9a⁴x³y³