f(x)=-1/(1+x²)=iA/(x+i)+iB/(x-i) as complex partial fractions.
So, iA(x-i)+iB(x+i)=-1,
iAx+A+iBx-B=-1, therefore, A+B=0, A-B=-1, and A=-½, B=½.
f(x)=f⁽⁰⁾(x)=(-i/2)/(x+i)+(i/2)/(x-i)=(i/2)(1/(x-i)-1/(x+i)).
Note that 1/i=-i. Also 1/i⁴=1=1/i⁰⇒1/i⁵=1/i¹=-1; 1/iⁿ=1/iⁿ⁻⁴.
f(x)=(i/2)(1/(x-i)-1/(x+i)); f⁽⁰⁾(0)=(i/2)(1/(-i)-1/i)=(i/2)(i+i)=-1
f⁽¹⁾(x)=(i/2)(-1/(x-i)²+1/(x+i)²); f⁽¹⁾(0)=(i/2)(-1/-1+1/-1)=0
f⁽²⁾(x)=(i/2)(2/(x-i)³-2/(x+i)³); f⁽²⁾(0)=i(1/i-(-1/i))=i(-i-i)=2
f⁽³⁾(x)=(i/2)(-3!/(x-i)⁴+3!/(x+i)⁴); f⁽³⁾(0)=(i/2)3!(-1+1)=0
f⁽⁴⁾(x)=(i/2)(4!/(x-i)⁵-4!/(x+i)⁵); f⁽⁴⁾(0)=(i/2)4!(i+i)=-24
...
f⁽ⁿ⁾(x)=(i/2)n!(1/(x-i)ⁿ⁺¹-1/(x+i)ⁿ⁺¹);
f⁽ⁿ⁾(0)=(i/2)n!(1/(-i)ⁿ⁺¹-1/iⁿ⁺¹).
General term for Maclaurin series is therefore (i/2)xⁿ(1/(-i)ⁿ⁺¹-1/iⁿ⁺¹).
For n=2r+1 for integer r≥0, the coefficient is zero, so the series is:
-1+x²-x⁴+x⁶-...