What this question means that if a number is say 98, it can be written as 10 x 9 + 8.

So, if a number appears as "xy", its value is actually 10x + y (because x is in the tens place and y is in the ones place).

Now, we have a two-digit number xy and its reversed yx where x > y.

The value of xy is 10x + y.
The value of yx is 10y + x.

Clearly xy is a bigger number than yx, so the square of xy is bigger than the square of yx.

Since the difference between the squares of the two numbers is 1584,
(10x + y)² - (10y + x)² = 1584
100x² + 10xy + 10xy + y² - 100y² - 10xy - 10xy - x² = 1584
99x² - 99y² = 1584
x² - y² = 16

Since the sum of the two numbers is 44 times the difference between the digits of the original number (bear in mind that x > y, so x - y will be the difference),
(10x + y) + (10y + x) = 44 (x - y)
11x + 11y = 44x - 44y
11y + 44y = 44x - 11x
55y = 33x
y = 0.6x

Then, our equation becomes
x² - (0.6x)² = 16
x² - 0.36x² = 16
0.64x² = 16
x² = 25
x = +- 5

Since x is positive number by default,
x = 5

Then,
y = 0.6 (5)
y = 3

So, the numbers are 53 and 35.