1.∵4f(x)/(1-x²)=d[f²(x)]
==>4f(x)/(1-x²)=2f(x)d[f(x)]
==>f(x){d[f(x)]-2/(1-x²)}=0
∴d[f(x)]-2/(1-x²)=0,或f(x)=0
（1）当d[f(x)]-2/(1-x²)=0时,
有d[f(x)]=2/(1-x²)
==>f(x)=∫2dx/(1-x²)
=∫[1/(1+x)+1/(1-x)]dx
=ln│1+x│+ln│1-x│+C (C是积分常数)
=ln│(1+x)/(1-x)│+C
∵f(0)=0 ==>C=0
∴f(x)=ln│(1+x)/(1-x)│
（2）显然f(x)=0是满足条件f(0)=0的解
综合（1）（2）知,f(x)=ln│(1+x)/(1-x)│,或f(x)=0
2.∵[ln(1+x+x²)+ln(1-x+x²)]/(xsinx)
=ln[(1+x+x²)(1-x+x²)]/(xsinx)
=ln(1+x²+x^4)/(xsinx)
=(x/sinx)*[ln(1+x²+x^4)/x²]
又lim(x->0)(x/sinx)=1/[lim(x->0)(sinx/x)]
=1 (∵lim(x->0)(sinx/x)=1)
lim(x->0)[ln(1+x²+x^4)/x²]
=lim(x->0)ln[(1+x²+x^4)^(1/x²)]
=ln{lim(x->0)[((1+x²+x^4)^(1/(x²+x^4)))(1+x²)]
=ln{lim(x->0)[e^(1+x²)]} (应用重要极限lim(x->0)[(1+x)^(1/x)]=e)
=lne
=1
∴原式=lim(x->0)(x/sinx)*lim(x->0)[ln(1+x²+x^4)/x²]
=1*1
=1