The smallest possible case is 11×11 = 121

Let the the maximal digit be x and the minimal digit be y

x ≧ y

The product is

(10x + y)(10y + x) < 1000
100xy+10x²+10y²+xy < 1000 
   10x²+101xy+10y² < 1000

Let y, the minimal digit, be small as possible, i.e., y=1

   10x²+101x+10 < 1000
   
   10x²+100x-990 < 0

The solution to that quadratic inequality is

-16.14 < x < 6.13

But since x is an integer, x = 1,2,3,4,5, or 6

61×16 = 976, and that is a 3-digit number

So

61×16, 51×15, 41×14, 31×13, 21×12, and 11×11

account for 6 cases.

Let y, be the next smallest digit possible i.e., y=2

   10x²+101xy+10y² < 1000

   10x²+101x(2)+10(2)² < 1000
   
          10x²+202x-40 < 1000

        10x²+202x-1040 < 0     

The solution to that quadratic inequality is

-24.45 < x < 4.25

Since x is an integer, x = 1,2,3, or 4

But 42×24 = 1008, is too large

However,
32×23 = 736 and 22×22 = 484 account for 2 more cases.

That's 8 cases.

The minimal digit y cannot be 3 since 33×33=1089,
which is not a 3 digit number.

So there are only 8 such 3-digit products.

Answer: fewer than 9.  In fact there are only these 8

61×16 = 976
51×15 = 765
41×14 = 574
31×13 = 403
21×12 = 252
11×11 = 121
22×22 = 484
23×32 = 736

Edwin