The problem is $y' = -\frac{1}{t^2} - \frac{1}{t}y + y^2;\ y_p = \frac{1}{t}$. My solution is
$$\begin{align} y = \frac{1}{t} + B &\implies y' = -\frac{1}{t^2} + B' \\ &\implies -\frac{1}{t^2} - \frac{1}{t}y + y^2 = -\frac{1}{t^2} + B' \\ &\implies -\frac{1}{t^2} - \frac{1}{t} \left(\frac{1}{t} + B \right) + \left(\frac{1}{t} + B\right)^2 = -\frac{1}{t^2} + B' \\ &\implies B' - \frac{1}{t}B = B^2 \\ &\implies L = B^{-1} \\ &\implies L' = -B^{-2} \left(B^2 + \frac{1}{t}B \right) \\ &\implies L' + \frac{1}{tB} = -1 \\ &\implies L' + \frac{1}{t}L = -1 \\ &\implies L_h = \frac{1}{t} \\ &\implies L = \frac{1}{t}\int\frac{-1}{\frac{1}{t}}dt \\ &\implies L = \frac{1}{t} \left(-\frac{1}{2}t^2 + C_{tentative} \right) \\ &\implies L = \frac{C - t^2}{2t} \\ &\implies B = \frac{2t}{C - t^2} \\ &\implies y = \frac{1}{t} + \frac{2t}{C - t^2} \end{align}$$
Although this solution does not appear equivalent to the answer given in my book, one or two people in chat reviewed this work and could not see anything incorrect. The issue is that the given particular solution, $y_p = \frac{1}{t}$, cannot be obtained by plugging any finite value into the $C$ in my general solution. We could say that the general solution yields the given particular solution when $C =$ infinity, more specifically, when $C =$ some $\aleph$ expression which makes the term $\frac{2t}{C - t^2}$ disappear regardless of $t$, but this feels a bit outside the box for a textbook problem. Is my general solution correct, and if so, how if at all can we derive the given particular solution from it?
This is a very common situation. A general solution for a nonlinear $n$-th order differential equation is a solution depending on $n$ independent parameters, but it does not necessarily give you all the solutions. As in this case, some may be obtained as limits where some parameters go to $\infty$. There are also cases where some particular solutions (singular solutions) are obtained as envelopes of the general solution.