mein Koeffizientenvergleich:

https://de.numberempire.com/equationsolver.php

F(x)=4/5x^5-4/3x^3+x=(a*8x^7-b*12x^5+c*6x^3-d*x)/(6x^2-1)

s(a,b,c,d)*(f(x)*f'(x)/f''(x))'=f(x)

(240*a*x^8+(-216*b-56*a)*x^6+(36*c+60*b)*x^4+(6*d-18*c)*x^2+d)/(6*x^2-1)^2=4*x^4-4*x^2+1

für "a" zb.:

a = (144*x^8+(216*b-192)*x^6+((-36*c)-60*b+88)*x^4+((-6*d)+18*c-16)*x^2-d+1)/(240*x^8-56*x^6)

für "b" zb.:

b = ((240*a-144)*x^8+(192-56*a)*x^6+(36*c-88)*x^4+(6*d-18*c+16)*x^2+d-1)/(216*x^6-60*x^4)

usw.

Ergebnisse des Koeffizientenvergleichs:

a=(144x^8)/(240x^8)=12/20

b=(192-56a)x^6/(216x^6)=132/180

c=-(60b-88)x^4/(36x^4)=22/18

d=1