I was given an example of the use of a function been Lipschitz to show that an ODE as a unique sloution
$$ \begin{cases} y'=y^2\\ y(0)=1\\ \end{cases} $$
now the solution is $y=\frac{1}{x+1}$ but how could we check that the function is Lipschitz without solving the ODE?
If we take $y=\frac{1}{x+1}$ and evaluate $y'=\frac{-1}{(x+1)^2}$ we can see that $|\frac{-1}{(x+1)^2}|=\frac{1}{(1+x)^2}\leq 1$
Solution of the ODE to be Lipschitz shouldn't be confused with the function defining the ODE to be Lipschitz.
For an ODE problem $y^\prime(x) = f(y,y(x))$, $y(x_0)=y_0$, the Cauchy-Lipschitz theorem states that if $f$ is uniformly Lipschitz continuous in $y$, then the ODE has a unique solution. That is the case of the ODE problem you consider.
To check this condition, you don't need to find the solution itself.
And here, the (maximal) solution is not Lipschitz continous as $\lim\limits_{x \to -1^+} y(x) = \infty$.