Investigate whether $ f $ meets Lipschitz continuity and whether it is uniform continuity 
Let $f: (-\infty,0]\rightarrow\mathbb R$ which is a function: $f(t)=\arcsin(e^{t})$ for $t\le0$.Investigate whether $ f $ meets Lipschitz continuity and whether it is uniform continuity

My try:
For $t\neq0$: $$f'(t)=\frac{e^{t}}{\sqrt(1-e^{2t})}$$$$f'(t)>1$$However $f'(t)$ has no upper limit so function $f'$ is not limited.We know that limited derivative $ \Leftrightarrow $ Lipschitz continuity for differentiable functions. So the first observation seems to be that $ f $ does not meet the Lipschitz continuity. But $f$ it is not differentiable in the whole domain so I don't know if I could use this observation.Moreover I know that function is uniform continuity when $$\forall_{\epsilon>0}\exists_{\delta>0} \forall_{x \in D}\forall_{y \in D}(|x-y|< \delta\Rightarrow |f(x)-f(y)|<\epsilon)$$Unfortunately I don't have idea how to show that $f$ (no)meets this condition.Can you get me some tips with it?

You have the derivative wrong: $f'(t)= e^t/\sqrt {1-e^{2t}}.$ You can use your Lipschitz-derivative criterion by noting that if $f$ were Lipschitz on $(-\infty,0],$ it would be Lipschitz on $(-\infty,0).$ On the latter interval $f$ is differentiable.
