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Does the question say anything about "extraneous solutions"? Have you learned of these kinds of solutions? I solved the problem you gave two different ways and I got y = -5 both times as well. And when I plugged it back in, it didn't work because it gave 0 as two of the three denominators. What the book might be giving as "Empty Set" means that there is no real answer to the problem if you discount the extraneous solution of y = -5.

But that's just me. I just finished Algebra 2 in school and I had one or two of these kinds of problems on the exam. And I got a really good grade, so I think that means I know what I'm doing. But I may be wrong. What you have up there isn't incorrect, technically, but they might be looking for a different solving method.

You did it correctly up to the y = -5. As Sioga says, that solution is extraneous. You'll probably have lots more cases like that. When you have polynomials in the denominator, square roots, or logarithms, etc., it's not uncommon to look like you're getting solutions at first and have to discard some or even all of them.

Easiest way to think of it is to just check: does it work in the original equation? And this doesn't. Byt what's really going on is when you're multiplying through by (y²-25) when you got rid of the denominator, you're masking the potential division by zero when y is -5 or +5. In other words, you're implicitly assuming that it doesn't happen, i.e., y is neither of those numbers.

So really the last steps should have a condition attached:
-y + 5 - 9y - 45 = 10,        as long as y ≠ ±5
-10y = 50,                    y ≠ ±5
y = -5,                     y ≠ ±5
Most of the time, people generally don't bother to write it, but it's still there. So the solution literally reads "y = -5, but y is not +5 or -5". And that's just as much a contradiction as "2 + y = y" ("2 = 0").