Rationalization is used to eliminate root radical in the denominator, for this both the numerator and denominator is multiplied by algebraic conjugate of denominator .For example if the denominator of an expression is \(\displaystyle{a}+\sqrt{{{b}}}\) then the algebraic conjugate is \(\displaystyle{a}-\sqrt{{{b}}}\). Now conjugate is multiplied to both numerator and denominator to get rationalized form of the expression.

In the given expression \(\displaystyle{\frac{{{3}\sqrt{{{y}}}}}{{{2}\sqrt{{{x}}}-{3}\sqrt{{{y}}}}}}\) the conjugate of denominator \(\displaystyle{\left({2}\sqrt{{{x}}}-{3}\sqrt{{{y}}}\right)}\) is \((2\sqrt{x}+3\sqrt{y})\)

To get the rationalize form multiply the numerator and denominator by \(\displaystyle{\left({2}\sqrt{{{x}}}+{3}\sqrt{{{y}}}\right)}\)

Use the algebraic identity \(\displaystyle{\left({a}+{b}\right)}{\left({a}-{b}\right)}={a}^{{2}}-{b}^{{2}}\) to simplify the denominator

Apply product rule of exponent \(\displaystyle{a}^{{m}}\cdot{a}^{{n}}={a}^{{{m}+{n}}}\)

\(\displaystyle{\frac{{{3}\sqrt{{{y}}}}}{{{2}\sqrt{{{x}}}-{3}\sqrt{{{y}}}}}}={\frac{{{3}\sqrt{{{y}}}{\left({2}\sqrt{{{x}}}+{3}\sqrt{{{y}}}\right)}}}{{{\left({2}\sqrt{{{x}}}-{3}\sqrt{{{y}}}\right)}{\left({2}\sqrt{{{x}}}+{3}\sqrt{{{y}}}\right)}}}}\)

\(\displaystyle={\frac{{{6}\sqrt{{{x}{y}}}+{9}{y}}}{{{4}{x}-{9}{y}}}}\)