Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\color{blue}{\dfrac{5x}{9y^5}}$
Use the rule $\frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c}=\frac{ad}{bc}$ to obtain: $\require{cancel} \\\frac{2x^2}{3y^4} \div \frac{6xy}{5} \\=\frac{2x^2}{3y^4} \cdot \frac{5}{6xy} \\=\frac{\cancel{2}\cancel{x^2}x}{3y^4} \cdot \frac{5}{\cancel{6}3\cancel{x}y} \\=\frac{x}{3y^4} \cdot \frac{5}{3y} \\=\frac{x(5)}{(3y^4)(3y)} \\=\color{blue}{\frac{5x}{9y^5}}$