Discussing $x' = \frac{1}{x+t}-1, x(1) = \xi$ and asymptotic stability

Question

I consider the differential equation

\begin{align} x' = \frac{1}{x+t}-1, \quad x(1) = \xi. \end{align}

For $\xi > -1$ (and with the help of $y(t) := x(t)+t$) one should find the unique maximal solution $\lambda_{\xi}: I_{\xi} \to \mathbb{R}$. I did that and found

\begin{align} \lambda_{\xi}(t) = \sqrt{\xi^2 + 2\xi + 2t -1}-t \end{align} with $I_{\xi} = \left\{t \in \mathbb{R} \mid t \geq \frac{1-\xi^2-2 \xi}{2}\right\}$.

Now it is asked to show, that $\lambda_0 : I_0 \to \mathbb{R}$ is an asymptotically stable solution of this differential equation. I really don't know how to do that.

My attempts have been:

1. Since the differential equation is of the form $x' = f(t,x)$ is would be enough to show that the solution is attractive, which would yield that it is also stable and hence asymptotically stable. But how do I show that the solution is attractive? (I know there is a definition for it but it struggle to apply it)

2. Since the differential equation is of the form $x' = f(t,x)$ I thought about looking at \begin{align} y'=f(t, y + \lambda_{\xi}(t))-f(t,\lambda_{\xi}(t)) \end{align} which has the trivial solution $y = 0$ and work from there on.

Would someone mind to show me how to use the definition or more general how to solve the question?