Point of inflection?

No , I believe it does....1st : x > 1 is required ; 2nd as x ---> ∞ this ----> 0 ; 3rd : x = 5 this function is > 0 while at x = 2 it is negative and at x = e it is 0....so it goes from - to 0 to + then back to 0 from above ====> inflection point exists

y = 10 ln (ln (x)) / ln (x)

y = 10 ln (ln (x)) (ln (x) )^(-1)
y' = 10 (ln (x) )^(-1) d/dx (ln (ln (x) ) + 10 ln (ln (x) ) d/dx ((ln (x))^(-1) )
y' = 10 (ln (x) )^(-1) 1 /(xln (x)) + 10 ln (ln (x) ) (-1) (ln (x) )^(-2) (1/x)
y' = 10 / (x ln^2 (x) ) -10 ln(ln(x) ) / (x ln^2 (x) )

y' = 10 x^(-1) (ln (x))^(-2) - 10 ln (ln (x) ) x^(-1) ( ln(x) )^(-2)
y'' = 10 x^(-1) d/dx ( (ln (x) )^(-2) + 10 (ln (x) )^(-2) d/dx (x^(-1) ) - 10 x^(-1) (ln (x))^(-2) d/dx (ln (ln (x) ) )
-10 ln (ln (x) ) d/dx ( x^(-1) ) (ln (x) )^(-2) - 10 x^(-1) ln (ln (x) ) (d/dx ( ln(x) )^(-2) )

y'' = 10 x^(-1) (-2) (ln(x) )^(-3) (1/x) + 10 (ln (x) )^(-2) (-1) x^(-2) - 10 x^(-1) (ln (x))^(-2) 1/(x ln(x))
- 10 ln (ln (x) ) (-1) x^(-2) (ln (x) )^(-2) -10 x^(-1) ln (ln (x) ) (-2) (ln(x))^(-3) (1/x)

y'' = -20 /(x^2 *ln^3 (x) ) -10 /(x^2 ln^2 (x) ) -10 / (x^2 * ln^3 (x) ) + 10 ln( ln(x) ) / (x^2 *ln^2 (x)) +20 ln (ln (x) ) / (x^2 *ln^3 (x) )
multiply and divide by x^2 ln^3(x)
set y'' = 0
y'' = (-20 -10 ln(x) -10 +10 ln (ln (x) ) + 20 ln (ln (x) ) ln^3(x) ) / (x^2 ln^3(x) ) = 0
y'' = -30 -10 ln(x) + 10 ln(ln (x)) + 20 ln (ln (x)) ln^3(x) = 0

x= 26.6971 is the point of inflection.
http://www.wolframalpha.com/input/?i=solve++-20+%2F(x%5E2+*ln%5E3+(x)+)+-10+%2F(x%5E2+ln%5E2+(x)+)+-10+%2F+(x%5E2+*+ln%5E3+(x)+)+%2B+10+ln(+ln(x)+)+%2F+(x%5E2+*ln%5E2+(x))+%2B20+ln+(ln+(x)+)+%2F+(x%5E2+*ln%5E3+(x)+)+%3D0

f(x) = (10ln(ln(x)))/(ln(x)

solve f"(x) = 0 --->ln(x) = -2 ---> no solution in the domain of f(x)