So far so good, you have

\(\displaystyle y=\frac{1}{\frac{1}{y_0}-t^2}\).

Now rearrange this a litlle by multiplying the top and the bottom by \(\displaystyle y_0\) to get

\(\displaystyle y=\frac{y_0}{1-y_0t^2}\).

Now your correct in saying that this exists provided the denominator is non-zero, or when \(\displaystyle y_0t^2 \not = 1\). But note that \(\displaystyle t^2\) is always greater than zero, hence when \(\displaystyle y_0\) is negative, it is impossible for \(\displaystyle t^2y_0\) to be positive, hence the equation exists for all \(\displaystyle y_0<0, t\). For \(\displaystyle y_0>0\) the denominator is equal to zero when \(\displaystyle t=+-\frac{1}{\sqrt{y_0}}\).

That agree with your book?

When

\(\displaystyle y_0 > 0\)

then your work and my work indicate that the solution exists for

\(\displaystyle t \ne \pm \frac{1}{\sqrt{y_0}}\)

the book says that the solution exists for

\(\displaystyle |t| < \frac{1}{\sqrt{y_0}}\)

If I set \(\displaystyle y_0 = 1\) and graph the corresponding formula \(\displaystyle y = \frac{1}{1-t^2}\), the graph does not exist at \(\displaystyle t = \pm 1\), but exists at all other values of t.

It would seem that the book is simply wrong? For one question I wouldn't be surprised (I find at least one or two mistakes in every text book), but there were three questions in a row that were differing from my answers like this and that usually means I am making the mistake rather than the book.