Provided 0 is a part of the domain of f(x)
f(0)=-f(-0)=>
f(0)=-f(0), (-0=0 obviously)
2f(0)=0
f(0)=0
This immediately implies if f(x) is an odd polynomial, 0 is a root, (x-0)=x is a factor (factor theorem), and hence f(x) is divisible by x.
To answer the f(x)=1/x question, it is a technicality not dealt with in high school but done in university, as if I define the domain of f(x) as all real numbers, then f(x)=1/x is not a function. This is because a function f(x) is has to map EVERY number in the domain to a number in the range of f(x), which in this case is all real numbers as well. Since f(0) is infinity (not a number) so then f(x)=1/x is not a function, so I cannot apply the definition of odd function to f(x)=1/x.
However, f(x)=1/x if x=/=0, while f(x)=0 if x=0 is a indeed an odd function. But again, this isn't dealt with in high school.