Answer:
[tex]\dfrac{6x^5y^2-16x^4y^3-6x^3y^4-9x^2+24xy+9y^2}{18y^2}[/tex]
Step-by-step explanation:
The applicable rule of exponents is ...
(a^b)(a^c) = a^(b+c)
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For this problem, it is convenient to factor out a factor that will eliminate fractions. Then the distributive property and the rules for adding fractions apply.
[tex]\left(\dfrac{2}{3}x^3-\dfrac{1}{y^2}\right)\left(\dfrac{1}{2}x^2-\dfrac{4}{3}xy-\dfrac{1}{2}y^2\right)\qquad\text{given}\\\\=\dfrac{1}{18y^2}(2x^3y^2-3)(3x^2-8xy-3y^2)\\\\=\dfrac{1}{18y^2}(2x^3y^2(3x^2-8xy-3y^2)-3(3x^2-8xy-3y^2))\\\\=\boxed{\dfrac{6x^5y^2-16x^4y^3-6x^3y^4-9x^2+24xy+9y^2}{18y^2}}[/tex]
