stuck with how this ordinary differential equation changes in new coordinates?

In the question it is assumed that Eq.$(1)$ : $$y'''+yy''=0 tag 1$$

is transformed into Eq.$(2)$ :

$$s_r s_{rrr}+r s_r^2 s_{rr}-3s_{rr}^2=0 tag 2$$

thanks to the change of variables : $quadbegin{cases} r=xy \ s=ln x end{cases}$

If this assumption is true any solution $y(x)$ of Eq.$(1)$ is transformed into $s(r)$ which is a solution of Eq.$(2)$.

For example try with $y(x)=x$ which is an obvious solution of $(1)$.

The change of variables transforms $y(x)=x$ into $s(r)=frac12ln(r)$.

$s_r=frac{1}{2r} quad;quad s_{rr}=-frac{1}{2r^2} quad;quad s_{rrr}=frac{1}{r^3}$

$s_r s_{rrr}+r s_r^2 s_{rr}-3s_{rr}^2=frac{1}{2r}(frac{1}{r^3})+r(frac{1}{2r})^2(-frac{1}{2r^2})-3(-frac{1}{2r^2})^2$

$s_r s_{rrr}+r s_r^2 s_{rr}-3s_{rr}^2=frac{1}{r^4}(frac12-frac{1}{8}-frac{3}{4})=-frac{3}{8r^4}neq 0$

The transformed function $s(r)$ is not solution of Eq.$(2)$.

This proves that the initial assumption is false. Possibly there is a typo in the question or a misunderstanding in the context.