Is $f:(-1,1)\to R$ defined as $f(x)= \frac{x}{1-\lvert x \rvert}$ and $f^{-1}(x)$ uniformly continuos? I think this fuction is uniformly continuos, but the inverse is not since the limit in the proximities to $\lvert x \rvert = 1$ the function has a vertical asymptote.
However I tried a lot of algebra to prove the inverse function is uniformly continuos, but with no luck. Any Hints? 
 A: Your function $f$ is not uniformly continuous. To see why, recall that a uniformly continuous function preserves Cauchy sequences but you can take a sequence $a_n\in(-1,1)$ such that $a_n\to 1$ but $f(a_n)\to\infty$, henece $f$ cannot be unifromly continuous.
A: This function cannot be uniformly continuous on the interval $(-1,1)$ due to the vertical asymptotes; you would have to cut back to $(-1+\epsilon, 1-\delta)$ in order to recover uniform continuity. See this post for some more tricks.
EDIT: To address the flip side of this problem, note that $f^{-1}:\mathbb{R}\rightarrow (-1,1)$ can be written out explictly:
$$
f^{-1}(x) = \left\{
        \begin{array}{ll}
            \frac{x}{1-x} & \quad x < 0 \\
            0 & \quad x=0 \\
            \frac{x}{1+x} & \quad x > 0
        \end{array}
    \right.
$$
On the intervals $(-\infty,-2]$ and $[2,\infty)$ note that $f^{-1}$ has a bounded derivative. Thus we know that $f^{-1}$ is uniformly continuous there. The interval $[-2,2]$ is compact and the function $f^{-1}$ is continuous on that interval, and thus is uniformly continuous on that interval. Finally, this post shows us that a function which is uniformly continuous on finitely many adjacent intervals is also uniformly continuous on their union.
