Correct. $\displaystyle\frac{{{3}{x}}}{{{5}{x}{{y}}^{{4}}}}\usepackage{cmbright}\Huge\$ and $\displaystyle\frac{{{6}{{x}}^{{2}}}}{{{15}{{x}}^{{2}}{y}}}\usepackage{cmbright}\Huge\$ have a common denominator of $\displaystyle{15}{{x}}^{{2}}{{y}}^{{4}}\usepackage{cmbright}\Huge\$ , so $\displaystyle\frac{{{3}{x}}}{{{5}{x}{{y}}^{{4}}}}+\frac{{{6}{{x}}^{{2}}}}{{{15}{{x}}^{{2}}{y}}}=\frac{{{\left({3}{x}\right)}{\left({3}{x}\right)}}}{{{15}{{x}}^{{2}}{{y}}^{{4}}}}+\frac{{{\left({{y}}^{{3}}\right)}{\left({6}{{x}}^{{2}}\right)}}}{{{15}{{x}}^{{2}}{{y}}^{{4}}}}=\frac{{{9}{{x}}^{{2}}+{6}{{x}}^{{2}}{{y}}^{{3}}}}{{{15}{{x}}^{{2}}{{y}}^{{4}}}}=\usepackage{cmbright}\Huge\$ $\displaystyle\frac{{{3}{{x}}^{{2}}{\left({3}+{2}{{y}}^{{3}}\right)}}}{{{15}{{x}}^{{2}}{{y}}^{{4}}}}=\frac{{{2}{{y}}^{{3}}+{3}}}{{{5}{{y}}^{{4}}}}\usepackage{cmbright}\Huge\$ .