Let $\alpha \in \mathbb{R}$ and $f$ be a real function defined by $f(u)=-ue^{\alpha u}\ln(\lvert u \rvert)$ if $u \neq 0$ and $f(0)=0$. Let $u_0 \in \mathbb{R}^+$ for the following problem:
\begin{equation} \begin{cases} u'(t)=f(u(t)), t\in \mathbb{R}\\ u(0)=0 \end{cases} \end{equation}
I'm trying to see under which conditions $f$ is locally Lipschitz. I tried the mean value theorem without much success.
Let us compute $f'(u)$ for $u > 0$: $$ f'(u) = -e^{\alpha u}\log(u)-\alpha u e^{\alpha u}\log(u) - e^{\alpha u} = -e^{\alpha u}(\log(u)+\alpha u \log(u) +1) $$
This is continuous on $(0,+\infty)$, so it is locally bounded on $(0,+\infty)$. Hence, $f$ is locally Lipschitz on $(0,+\infty)$.
However, $$ \lim_{u \to 0⁺} f'(u) = +\infty, $$ hence $f$ is not locally Lipschity on $\mathbb{R}$.
If your initial condition $u_0$ is strictly positive, there is local (in time) existence and uniqueness to the Cauchy problem.