If h(x) = |x+3| — |x— 3| for all real x, then for how many integer values of x is the inequality |h(x)| < 6 satisfied?

The question asks us the integer values of x for which, the inequality
$$\left|\left|x+3\left|-\right|x-3\right|\right|<6$$ is satisfied. 
We can also interpret the given inequality as, 
$$-6<\left|x+3\left|-\right|x-3\right|<6$$

Now we see that there are two key points to take note of, where the modulus shows it's effects. Those are at x=-3 and x=3
We need to check the value of the middle term for the three cases of x<-3, -3<x<3 and 3<x

For x< -3, 
We see that the first and second modulus terms will have negative values inside them, and will thus reverse the sign. 
We can take a value as an example to see the result. Let's take x= -10
$$\left|-10+3\right|-\left|-10-3\right|$$
$$\left|-7\right|-\left|-13\right|$$
$$7-13\ =\ -6$$
For any negative value of x < -3, the answer will be -6. 
This does not satisfy our condition as we need the resulting values bigger than -6.

For x> 3,
We see that both the terms in the modulus will be positive and hence the modulus will not change any signs.
$$\left(x+3\right)-\left(x-3\right)$$
$$x+3-x+3\ =\ 6$$
For any value of x> 3, the answer would be 6. 
This again does not satisfy our condition and thus is not valid. 

For -3< x< 3, 
For this range, the first modulus term will have positive value inside while the second modulus modulus will have a negative value inside. 
$$\left(x+3\right)-\left(-\left(x-3\right)\right)$$
$$\left(x+3\right)-\left(-x+3\right)$$
$$\left(x+3\right)+x-3$$
$$2x$$
And since the values of x lies between -3 and 3, the values of 2x lies between -6 and 6. 
And thus this range satisfies the inequality.

The integers that lie in this range are -2, -1, 0, 1 and 2.
So a total of 5 integers satisfy this inequality.