I'm really confused , should I use the rule which solves it as
$$\ln\left(\frac 1{a^2}\right) = \ln (1) - \ln(a^2)$$
Or should I first put $\dfrac 1{a^2} = a^{-2}$ and then take the log which gives me the result $\ln(a^{-2})$ ?
In the second approach the term $\ln 1$ is not present; am I using that log rule incorrectly?
They are equivalent: $$\ln\left(\frac 1{a^2}\right) = \underbrace{\ln 1}_{=\,0} - \ln (a^2) = 0 - \ln(a^2) = -\ln(a^2) = \ln \left((a^2)^{-1}\right) = \ln\left(a^{-2}\right) = \ln\left(\frac 1{a^2}\right)\tag 1$$
Note: From $-\ln(a^2)$ we can further reduce this as $$-\ln(a^2) = -2\ln a\tag 2$$
The fact that $\ln (1) = 0$ (as used in (1)) can also be verified by raising each side as an exponent of $e$: $$e^{\ln 1} = 1 = e^0.$$ Also good to remember is $$\ln(x^y) = y\ln(x),\,\tag{ as used in (2)}$$