If $$y = {\left( {1 + \frac{1}{x}} \right)^x}$$   then $$\frac{{2\sqrt {{y_2}\left( 2 \right) + \frac{1}{8}} }}{{\left( {\log \frac{3}{2} - \frac{1}{3}} \right)}}$$    is equal to -

Solution :
Let $$y = {\left( {1 + \frac{1}{x}} \right)^x}$$
Taking logarithm of both sides, we get
$$\eqalign{ & \log \,y = x\left[ {\log \left( {1 + \frac{1}{x}} \right)} \right] \cr & \Rightarrow \frac{1}{y}{y_1}\left( x \right) = \frac{{{x^2}}}{{x + 1}}\left( { - \frac{1}{{{x^2}}}} \right) + \log \left( {1 + \frac{1}{x}} \right) \cr & = - \frac{1}{{x + 1}} + \log \left( {1 + \frac{1}{x}} \right).....(1) \cr & {\text{Since,}}\,y\left( 2 \right) = {\left( {1 + \frac{1}{2}} \right)^2} = \frac{9}{4} \cr & {\text{So, }}{y_1}\left( 2 \right) = \left( {\frac{9}{4}} \right)\left( { - \frac{1}{3} + \log \frac{3}{2}} \right) \cr} $$
Again differentiate equation (1) w.r.t. $$\left( x \right),$$ we get
$$\frac{{y\left( x \right){y_2}\left( x \right) - {{\left[ {{y_1}\left( x \right)} \right]}^2}}}{{{{\left( {y\left( x \right)} \right)}^2}}} = \frac{1}{{{{\left( {1 + x} \right)}^2}}} - \frac{1}{{x\left( {x + 1} \right)}}$$
By putting $$x = 2,$$  we get
$$\frac{{y\left( 2 \right){y_2}\left( 2 \right) - {{\left( {{y_1}\left( 2 \right)} \right)}^2}}}{{{{\left( {y\left( 2 \right)} \right)}^2}}} = \frac{{ - 1}}{{18}}$$
Now, put value of $$y\left( 2 \right)$$  and $${y_1}\left( 2 \right)$$
$$\eqalign{ & \Rightarrow {y_2}\left( 2 \right) = \left( {\frac{9}{4}} \right){\left( { - \frac{1}{3} + \log \frac{3}{2}} \right)^2} - \frac{1}{8} \cr & \Rightarrow {\left( {{y_2}\left( 2 \right) + \frac{1}{8}} \right)^4} = 9{\left( {\log \frac{3}{2} - \frac{1}{3}} \right)^2} \cr & \Rightarrow {\text{ Required expression }} = 3 \cr} $$