Calculus: Early Transcendentals, Chapter 3, 3.6, Section 3.6, Problem 24

You need to find the first derivative of the function, using the quotient rule, such that:
y' = ((ln x)'*x^2 - ln x*(x^2)')/((x^2)^2)
y' = (x^2/x - 2x*lnx)/(x^4)
y' = (x - 2x*lnx)/(x^4)
You need to evaluate the second derivative, differentiating the first derivative, with respect to x, such that:
y'' = ((x - 2x*lnx)'(x^4) - (x - 2x*lnx)(x^4)')/((x^4)^2)
y'' = ((1 - 2*lnx - (2x)/x)(x^4) - 4x^3(x - 2x*lnx))/(x^8)
y'' = ((1 - 2*lnx - 2)(x^4) - 4x^3(x - 2x*lnx))/(x^8)
Factoring out x^3, yields:
y'' = x^3*(x*(1 - 2*lnx - 2) - 4(x - 2x*lnx))/(x^8)
Reducing like terms, yields:
y'' = ((1 - 2*lnx - 2) - 4(x - 2x*lnx))/(x^5) = (6lnx - 5)/(x^5)
Hence, evaluating the first and the second derivatives, yields y' = (x - 2x*lnx)/(x^4) and y'' = (6lnx - 5)/(x^5)