CertainPerformance has explained why your code wasn't working.
Here is another implementation, one I find reasonably simple:
const reverseNum = (n) =>
n < 10
? String(n)
: String (n % 10) + reverseNum (Math .floor (n / 10))
console .log (reverseNum (8675309))
Note that this returns a String rather than a Number. If we wanted to, we could make this a private function and make a public function one which called this and converted the result back into a number. But that would have the weird effect that reversing, say, 1000 would yield 1, since 0001 is simplified to 1. And that would mean that when you reverse again, you don't get anything like your original value. So I choose to keep with a String.
Of course if we're going to do String reversal, perhaps we're better off just using a String reversal function in the first place:
const reverseStr = (s) =>
s.length == 0
? ''
: reverseStr (s .slice (1)) + s [0]
const reverseNum = (n) =>
reverseStr (String(n))
console .log (reverseNum (8675309))
Or if we weren't interested in doing this recursively, we could just write the more common string reversal function:
const reverseStr = (s) =>
s .split ('') .reverse () .join ('')
const reverseNum = (n) =>
reverseStr (String (n))
console .log (reverseNum (8675309))