∵函数f（x）=ax^3+bx^2+cx+d的图像关于原点对称
∴ f(x)为奇函数
则f(-x)=-f(x)
即-ax^3 +bx² -cx +d = -(ax^3+bx^2+cx+d) = -ax^3-bx^2 - cx- d
比较系数可得：b = 0,d = 0
∴f(x) = ax^3 +cx
f'(x) = 3ax² + c
∵x=1时f(x)去最小值-2/3
∴f(1)=a+c= -2/3
f'(1)=3a+c=0
解得：a=1/3,c=-1
即f(x) = x^3/3 - x