Let the digits be a1,a2,a3,… d4, using standard algebraic chess notation (i.e. letters = columns, numbers = rows, a1 = lower left).
Define a low digit = 1-8, high digit = 9-16. The cells, b1,a2,b2,c2,d2,c3 are all low. The top row contains two low numbers otherwise d4 > 16. This accounts for all eight low numbers and therefore a3 is high. Ergo, a4=1 to avoid a1>16. Now consider the chain c4 < c1 < d1 < d4. The minimum differences are c1-c4 >= 6, d1-c1 >= 2, d4-d1 >= 2. We soon find a2,c2,c3 must be some permutation of 2,3,4. Therefore c4 is at least 5 and d4 is at least 15.
Since 13 is prime, it must go in a3 or d1. But if a3 = 13 then a1,d3,d4 are all 15 or greater, contradiction. Therefore d1=13, c1=11, c4=5. We have a 2,3 pair in c2,c3 so a2 = 4. Therefore d2=8 and d3=16. Since b2/c2 = 2 we have 4*6/3=8 in row 3 and the rest is easy