You may have thought that you could rewrite the numerator as $\displaystyle\sqrt{{{3}+{b}}}\usepackage{cmbright}\Huge\$ , then proceeded to rationalize the denominator. You may have been thinking of $\displaystyle\sqrt{{3}}\cdot\sqrt{{b}}=\sqrt{{{3}{b}}}\usepackage{cmbright}\Huge\$ , but there is no similar rule for addition. The first two steps are:

$\displaystyle\frac{{{3}+\sqrt{{b}}}}{{{3}\sqrt{{{b}}}}}\cdot\frac{\sqrt{{{b}}}}{\sqrt{{{b}}}}=\frac{{{3}\sqrt{{{b}}}+\sqrt{{{{b}}^{{2}}}}}}{{{3}\sqrt{{{{b}}^{{2}}}}}}\usepackage{cmbright}\Huge\$