Terminating, Purely Periodic, or Eventually Periodic?

JavaScript (ES6), 70 .. 68 53 bytes

f=(a,b,s=[],x)=>a?(s[a]^=a)?f(a*10%b,b,s,x||a):x==a:0

Returns 0 for terminating, true for purely periodic and false for eventually periodic.

How it works

What we're doing here is actually simulating a division by hand:

1. a?...:0 - If the numerator is zero, we stop here and return 0. The sequence is terminating.
2. (s[a]^=a)?...:x==a - If we've already encountered this numerator before, it means that the sequence is periodic and is going to repeat forever. We stop here and return either true if a is equal to the first value x of the sequence (purely periodic) or false if it's not (eventually periodic).
3. f(a*10%b,b,s,x||a) - Else, we multiply the numerator a by 10. We compute the remainder of the division by the denominator b. And we repeat the process by using this remainder as the new numerator. (We also pass a as the first value of the sequence if it's not already stored in x.)

Example

• Blue: numerator = 1
• Green: denominator = 7
• Red: multiplications by 10
• Black: remainders
• Gray: quotient digits (we don't really care about them here, and the above code is not computing them at all)

Python, 62 61 59 bytes

f=lambda n,d,r=[0,0]:(r[:3]+r).count(n)or f(10*n%d,d,r+[n])

Prints 1 for eventually periodic, 2 for purely periodic, and 4 for terminating.

Verify all test cases on repl.it.

Jelly, 10 bytes

:gÆfḍ⁵ṢQ¬Ḅ

Accepts denominator and numerator (in that order) as arguments. Returns 0 for terminating, 1 for purely periodic, and 2 for eventually periodic. Try it online! or verify all test cases.

How it works

:gÆfḍ⁵ṢQ¬Ḅ  Main link. Arguments: d (denominator), n (numerator)

 g          Compute the GCD of d and n.
:           Divide d by the GCD, yielding the denominator of the simplified form.
  Æf        Yield all prime factors of the previous result.
    ḍ⁵      Test 10 for divisibility by each prime factor.
            This yields 1 for 2 and 5, 0 for all other primes.
      Ṣ     Sort the resulting Booleans.
       Q    Unique; deduplicate the sorted Booleans.
        ¬   Logical NOT; replace 0 with 1 and vice versa to yield one of the
            following arrays.
              [    ]  <- no prime factors (denominator 1)
              [   0]  <- only 2 and 5
              [1   ]  <- neither 2 nor 5
              [1, 0]  <- mixed
         Ḅ  Unbinary; convert from base 2 to integer.
            This maps [] and [0] to 0, [1] to 1, and [1, 0] to 2.