The basic idea of reducing expressions like this is relatively simple, but doing so by hand takes quite a bit of effort. In principle one could start working directly from the given form, and with some experience it would not matter much. But if you have to do this by hand, it is probably better to make things as simple as possible first. I also need to point out that there is not a unique simplification.
The first step is write the expression in a more basic form by separating all products inside the logarithms by using the relation:
$$\ln a^m b^n = m \ln a + n \ln b$$
on the whole numbers $m$ and $n$ and positive arguments $a$ and $b$. In this case you only need to consider the cases with $m,n \in \{-2,-1,0,1,2\}$ and the only arguments in the logarithms that remain are $\{x,z,y+z,2 y +z,x-y^2\}$.
The expression would look something like
$$
10 \ln^2(y+z) + 3 \left[ \ln z + \ln (2 y +z) \right]^2 + \dots
$$
The first and simplest approach to transform the expression into something with dimensionless arguments only, is based on using the most basic unit(s) that exist. Since in this case $x$ has dimension 2 and $y$,$z$ have dimension 1 the later are more basic. So for simplicity I will choose $z$ because that also appears as an argument in the logarithms. The only thing we have to do now, is consider term by term and improve the expression.
The first term is $\ln^2(y+z)$ which is of dimension 1, but we can replace it by $(\ln \frac{y+z}{z} + \ln z)^2$. For every term $\ln [\dots]$ we have in the expression we replace it by $\ln(\frac{[\dots]}{z^n}) + n \ln z$ where $n$ is the dimension, positive or negative, of the argument $[\dots]$. If we do this for every term in the given expression and expand the result we are left with dimensionless arguments in all logarithms. The only other terms that could have survived would be of the form $\ln^n z$ and if that is the case (and we did not make a mistake) the original expression cannot be written in dimensionless terms only.
For this the initial reduction in basic terms is strictly speaking not required, but some other mixed terms might remain that require a bit of extra attention.
An alternative approach which cost more effort would be, again starting from the reduced form in the 5 basic elemental argument, look for patterns. If done correctly, you would have found something like
$$
3 \ln^2(2 y + z) + 6 \ln(2 y + z) \ln z - 12 \ln(2 y + z) \ln (y+z) + \dots
$$
These are all the terms that contain a factor $\ln(2 y + z)$. It is now important to recognise the patterns and identify the mixed terms which allow us to rewrite this as:
$$
6 \left[ \ln(2 y + z) - \ln (y+z) \right]^2 - 3 \left[\ln(2 y + z) - \ln z\right]^2 - 6 \ln^2 (y+z) + 3 \ln^2 z \dots
$$
and this gives us
$$
6 \ln^2 \left(\frac{2 y + z}{y+z}\right) - 3 \ln^2\left( \frac{2 y + z}{z} \right) - 6 \ln^2 (y+z) + 3 \ln^2 z + \dots
$$
In so doing we added some 2 extra terms, but more importantly we eliminated al logarithmic terms with non-dimensionless arguments that contained a factor $(2y+z)$. A different way in which this could have been achieved would have been using
$$
3 \left[ \ln(2 y + z) - 2 \ln (y+z) + \ln z \right]^2 + \dots = 3 \ln^2 \left( \frac{z(2 y+z)}{(y+z)^2} \right) + \dots
$$
The steps can be repeated with the other basic arguments in order to remove all other powers and products of logarithmic terms.
In any case, doing this by hand is a bit cumbersome and time consuming, but the process can be automated and can easily be done in a program like Mathematica. In the case of the expression you have given, however, there will be a term $\frac{10}{3} \ln z$ left that cannot be made dimensionless. So possibly it contains a typo or a mistake in the derivation that gave this expression.
