I have been stressing over this one cryptarithm. Can you help me solving it?

This problem is a very interesting one!
Here is the solution, step by step:

If "t" in the multiplier times "hi" in the multiplicand yields "at" in the first partial product; also "n" in the multiplier times "hi" in the multiplicand yields "fin" in the second partial product, this suggests that i = 1 or otherwise i = 6 and "t" and "n" are even numbers. Let´s first try i = 1. Substituting "i" for "1" the set-up would become:

Easy to see that "o" should equal 2 because it results from "1" plus a "carry 1" from the sum a + n = o + 10. So we come to:

In order to get "n" times h1 = f1n we must have n = 3 and h = 7 or otherwise n = 7 and h = 3. And in both cases "f" should equal 2. But this is impossible because we have already determined o = 2. Hence we must go back to where we started and now make i = 6 as far as i = 1 has revealed impossible. Then we start again with:

We also know that "t" and "n" are even numbers. Easy to deduct that
o = 7 (it results from "6" plus a "carry 1" from the product a + n = o + 10) Hence:

In the column printed with characters in bold we see that a + n = 17. For this to be true there are only two alternatives: 9 + 8 = 17 or 8 + 9 = 17. But we also know thar "n" is an even number, so that makes
a = 9 and n = 8. Substituting we get:

In the first partial product we see that "t" times "h6" equals "9t". Hence
h = 4 and t = 2 are the only numbers that fit. Substituting we get:

It is evident that f = 3 leading us to the final solution:

The section "EXAMPLES WORKED OUT IN DETAIL BY MASTER PUZZLISTS" contains 6 puzzles solved step-by-step by renowned puzzlists, where you can learn many basic tips for problem-solving.