does the uniform continuity of $f$ implies uniform continuity of $f^2$ on $\mathbb{R}$? my question is if $f:\mathbb{R}\rightarrow\mathbb{R}$ is uniformly continuous, does it implies that $f^2$ is so?and in general even or odd power of that function?
 A: By $f^2(x)$, do you mean $(f(x))^2$, or $f(f(x))$?  (The notation $f^2(x)$ can mean both of these; which is standard depends on the area of mathematics you work in.)
If you mean $(f(x))^2$, then as other answers have pointed out, this may not be uniformly continuous.
However, if you mean $f(f(x))$, then yes, this is uniformly continuous whenever f is.
Proof.  Given $\epsilon > 0$, we need some $\delta > 0$ such that whenever $|x - y| < \delta$, $|f(f(x)) - f(f(y))| < \epsilon$.
We know, by uniform continuity of $f$ applied to $\epsilon$, that there’s some $\epsilon_1$ such that whenever $|x - y| < \epsilon_1$, $|f(x) - f(y)| < \epsilon$.  By the uniform continuity of $f$ again, applied to $\epsilon_1$, there’s $\delta$ such that whenever $|x - y| < \delta$, $|f(x) - f(y)| < \epsilon_1$.
So putting these together, if $|x - y| < \delta$, then $|f(x) - f(y)| < \epsilon_1$, and so $|f(f(x)) - f(f(y))| < \epsilon$; so $\delta$ is as required.  $\square$
A: As mentioned, this is not true.  However, a related statement is true: If $f$ is uniformly continuous and bounded, then $f^2$ is uniformly continuous (and bounded).  (Hint to prove this: $|f(x)^2 - f(y)^2| = |f(x)+f(y)| |f(x)-f(y)|$).  Indeed, if $f,g$ are both uniformly continuous and bounded, then so is $fg$.  Thus any finite product of uniformly continuous bounded functions is another such; in particular, if $f$ is, then so is $f^n$ for any $n$.
A fancy version of this statement is that the set $C_u(\mathbb{R})$ of all uniformly continuous bounded functions on $\mathbb{R}$ is a $C^*$-algebra.
A: No. For example, $f(x)=x$ is uniformly continuous, but $f(x)=x^2$ is not, and neither for that matter is $f(x)=x^n$ for any $n>1$.
A: No. If $f(x)= x$ then $f^2(x)$ is not unifomly continuous. A statement like that will only hold on compact sets.  
