Let y=x3/(x2-1)=x+x/(x2-1).
Let x/(x2-1)=A/(x-1)+B/(x+1), then Ax+A+Bx-B=x; A-B=0 and A+B=1, so A=B=½.
y=x+½(x-1)-1+½(x-1)-1); y'=y(1)=1-½(x-1)-2-½(x+1)-2; y(2)=½[2(x-1)-3+2(x+1)3]; etc.
y(n)=½(-1)n[n!(x-1)-(n+1)+n!(x+1)-(n+1)]. But y'=1-½-½=0.
When x=0, y(n)=½(-1)n[n!(-1)-n-1+n!(1)-n-1].
When n>1 is odd: y(n)=-½[n!+n!]=-n!; when n is even: y(n)=½[-n!+n!]=0 QED