$L-f(h)=ah^6+bh^9$
$L-f(h/2)=a(h/2)^6+b(h/2)^9$
Now cancel the $a$ term between the two
$[L-f(h)]-64[L-f(h/2)]=bh^9-64b(h/2)^9$
$64f(h/2)-f(h)-63L=7b/8h^9$
$L=(64f(h/2)-f(h))/63-b/72h^9$
Hence
the best approximation for $L$ from this step of Richardson is $(64f(h/2)-f(h))/63$.
You can see that if the first exponent is $n$ (in place of $6$) then $\displaystyle L\approx {2^nf(h/2)-f(h) \over 2^n-1}$