This follows from the quotient rule:
[tex]y=\dfrac{1+\ln t}t\implies\dfrac{\mathrm dy}{\mathrm dt}=\dfrac{t\left(\frac1t\right)-(1+\ln t)}{t^2}[/tex]
and simplifying a bit you end up with
[tex]\dfrac{\mathrm dy}{\mathrm dt}=-\dfrac{\ln t}{t^2}[/tex]
(note that [tex]y(t)[/tex] is only defined for [tex]t>0[/tex], which lets us simplify [tex]t\left(\dfrac1t\right)=1[/tex])
