By the quotient rule, the first derivative is #f'(x)=(x^2 * 1/x - ln(x) * 2x)/(x^4)=(x-2xln(x))/(x^4)=(1-2ln(x))/(x^3)#. Use the quotient rule again to find the second derivative:

#f''(x)=(x^3 * -2/x - (1-2ln(x)) * 3x^2)/(x^6)#

#=(-2x^2-3x^2+6x^2ln(x))/(x^6)=(6ln(x)-5)/(x^4)#

For extra interest, the fact that #f'(x)=(1-2ln(x))/(x^3)# implies that #f# is increasing for #0 < x < e^{1/2} approx 1.65# and decreasing for #x > e^(1/2)#, with a local maximum value of #f(e^{1/2})=(1/2)/((e^{1/2})^2)=1/(2e) approx 0.18# at #x=e^{1/2} approx 1.65#. (Make sure you check all this!)

The fact that #f''(x)=(6ln(x)-5)/(x^4)# implies that #f# is concave down for #0 < x < e^{5/6} approx 2.30# and concave up for #x > e^{5/6}#, with an inflection point at #(x,y)=(e^{5/6},f(e^{5/6}))=(e^{5/6},(5/6)/((e^{5/6})^2))=(e^{5/6},5/(6e^{5/3})) approx (2.30,0.16)#. (Make sure you check all this!)

Here's the graph to allow you to see these features.

graph{ln(x)/(x^2) [-0.562, 4.438, -0.77, 1.73]}