Correct. Using a common denominator of $\displaystyle{40}{{x}}^{{3}}{y}\usepackage{cmbright}\Huge\$ , you find that $\displaystyle\frac{{{5}{{x}}^{{2}}}}{{{8}{x}{y}}}-\frac{{{3}{y}}}{{{10}{{x}}^{{3}}}}=\frac{{{25}{{x}}^{{4}}-{12}{{y}}^{{2}}}}{{{40}{{x}}^{{3}}{y}}}\usepackage{cmbright}\Huge\$ .