Let g(x) be differenciable at c. And g(c)=0.

Then we can define h(x) = 1/g(x) for x!=c on some open interval containing c. (See proof below).

Proof: Since g(c) is not zero there is an interval containing c so that g(x)!=0 except at c. The reason is because g(x) is differenciable on I so it is continous on I. Thus, |g(x)-g(c)| < e. Thus, g(c) - e< g(x) < g(c) + e. If g(c) >0 then choose e small enough so that g(c) - e>0. And if g(c) < 0 choose e small enough so that g(c) + e<0. On the interval |x-c|< d so c-d<x<c+d. So choose (c-d,c+d) as your interval.

Q.E.D.

This tells us that 1/g(x) is defined on an open interval containing c except possible at c.

We can therefore define h(x) = (1/x) o (g(x)) and use the chain rule.