Simplify the first and last terms using $\displaystyle\sqrt{{{a}{b}}}=\sqrt{{a}}\cdot\sqrt{{b}}\usepackage{cmbright}\Huge\$ :

$\displaystyle{16}\sqrt{{{{x}}^{{3}}{y}}}+{\sqrt[{{4}}]{{{{x}}^{{3}}{y}}}}+\sqrt{{{9}{{x}}^{{3}}{y}}}={16}{\left|{x}\right|}\sqrt{{{x}{y}}}+{\sqrt[{{4}}]{{{{x}}^{{3}}{y}}}}+{3}{\left|{x}\right|}\sqrt{{{x}{y}}}\usepackage{cmbright}\Huge\$

The first and last terms are like terms and can be combined:

$\displaystyle{16}{\left|{x}\right|}\sqrt{{{x}{y}}}+{3}{\left|{x}\right|}\sqrt{{{x}{y}}}={\left({16}+{3}\right)}{\left|{x}\right|}\sqrt{{{x}{y}}}={19}{\left|{x}\right|}\sqrt{{{x}{y}}}\usepackage{cmbright}\Huge\$

So the entire expression equals $\displaystyle{19}{\left|{x}\right|}\sqrt{{{x}{y}}}+{\sqrt[{{4}}]{{{{x}}^{{3}}{y}}}}\usepackage{cmbright}\Huge\$ . Because the indices are not the same, this expression cannot be further simplified.