Now see if you can justify that guess by doing as the hint suggests and using the definition of the derivative. First, note that . Then , so that .
The idea now is to find a bound for that last integral. The other factor will then ensure that the expression on the left side of the inequality goes to 0 as , which will show that is differentiable at .
To get a bound on that integral, let d be the distance from to the interval [0,1] (you are told that , so you know that d>0), and choose z close enough to to ensure that the distance from z to [0,1] is greater than d/2. That will give you a lower bound on the denominator of the integrand; and the fact that h is continuous means that it is bounded on [0,1], so that gives you an upper bound for the numerator.