Correct. You recognized that $\displaystyle\frac{{{x}+{3}}}{{{{x}}^{{2}}+{3}{x}}}=\frac{{{x}+{3}}}{{{x}{\left({x}+{3}\right)}}}\usepackage{cmbright}\Huge\$ .

$\displaystyle\frac{{{x}+{3}}}{{{x}+{3}}}={1}\usepackage{cmbright}\Huge\$ , so $\displaystyle\frac{{{x}+{3}}}{{{x}{\left({x}+{3}\right)}}}={1}\cdot{\left(\frac{{1}}{{x}}\right)}=\frac{{1}}{{x}}\usepackage{cmbright}\Huge\$ .