简单计算一下即可，答案如图所示
设u=x+y，则y=f(x+y)=f(u) y'=dy/dx=df/dudu/dx=f'(u)u'=f'(u)(1+y')。所以有
y'=f'(x+y)(1+y') 所以：y'=f'(x+y)/(1-f'(x+y))
y"=d(f'(x+y)(1+y'))/dx
=f"(x+y)(1+y')²+f'(x+y)y"
y"=(f"(x+y)(1+y')²)/(1-f'(x+y))
=f"(x+y)/(1-f'(x+y))³
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