Let me put them as a,b,c for our notation. So a,b,c are different,non-zero single digit numbers
now,
ab * ba =cbca
and
a(a+c) = b

a(a+c) = b
we know that b is a single digit
product of two single digits giving a single digit
how many cases:
1* (1+{2,3,4,5,6,7,8,9} )

Hence a=1 - Let us keep this for some time..

another possibility..
2*(2+1)
because c cannot be 2 if a=2. [different digits]
c cannot be 3 because b will become 10.
so a=2, c=1 -> let us keep this also aside

next:
3*(3+0)
this is not possible because c cannot be 0. And if c is greater than 0 them b is more than 10.

So for any value of "a" greater than 2, we will not have a feasible solution.

So now a has to be either 1 or 2.
when a=2, c=1, b= 6
coming to our other equation:
ab * ba =cbca
26*62=1612

This is satisfied.

Let us see what if a=1?
Look at the last digit of the product -> b*a is yielding last digit as "a"
if a=1, then last digit will be "b" and not "a"

Hence a cannot equal 1.

hence ab = 26

Here's how I did it. It's a bit unconventional, so looking fwd to better explanations!

Rephrase problem as :

=>xy
* yx
____
zyzx

We also have x(x+z) = y

Start with this info to narrow down possibilities.

1. Since x,y and z are all different non-zero single digit integers, we can infer that y is a multiple of x, and since y cannot be equal to x, it must be 2x, 3x, etc.

2. Since minimum value of y is 2x and maximum value is 9, we can also infer that x < 5 since 2*5 = 10 which violates our single-digit constraint for y.

So this leaves our possible values of x and y as :
x = 1, y = 2,3,...
x = 2, y = 4,6,8
x= 3, y = 6,9
x = 4, y = 8

Now we add our final constraint which is visible from the product. The units digit of the product is "x" so this means that the product of x*y must have a units digit of x.

x= 1 will give units digit of 2,3,... but not 1.
x = 2 gives us a units digit of 2 when y = 6

Test this to see if it satisfies our condition: 26*62 = 1612 => z = 1 which satisfies x(x+z) = y

Pick 26.