[tex] x^{\frac{1}{2}} = \sqrt[2]{x} [/tex] and [tex] 3^{-\frac{1}{2}} = \frac{1}{3^{\frac{1}{2}} } = \frac{1}{\sqrt[2]{3}} [/tex] so
[tex] 3^{-\frac{1}{2}} x^{\frac{1}{2}} = \frac{1}{\sqrt{3}} \cdot \sqrt{x} = \frac{\sqrt{x}}{\sqrt{3}} [/tex]
and you rationalize denominator by multiplying numerator and denominatr by [tex] \sqrt{3} [/tex] so that gives--
[tex] \frac{\sqrt{x}}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{3x}}{3} [/tex]
Your answer is [tex] \frac{\sqrt{3x}}{3} [/tex]
