Smoothness of $f(x)/(1+|f(x)|)$ where $f \in C^1(E)$ for $E$ an open subset of $\mathbb{R}^n$ (a) Show that if $E$ is an open subset of $\mathbb{R}$ and $f \in C^1(E)$ then the function
$$F(x) = \frac{f(x)}{1+|f(x)|}$$
satisfies $F \in C^1(E)$.
(b) Extend the results of part (a) to $f \in C^1(E)$ for $E$ an open subset of $\mathbb{R}^n$.
Hint part (a): Show that if $f(x) \neq 0$ at $x \in E$ then $$F'(x)=\frac{f'(x)}{(1+|f(x)|)^2}$$ and that for $x_0 \in E, f(x_0)=0$ then $F'(x_0)=f'(x_0)$ and that $lim_{x \rightarrow x_0} F'(x)=F'(x_0)$
Comments: Showed part (a), but I can not generalize to the $\mathbb{R}^n$.
 A: Both (a) and (b) are best dealt with simultaneously, by writing $F$ as the composition of $f$ with the function 
$$\phi(t) = \frac{t}{1+|t|} , \quad t\in\mathbb R$$ 
Note that $\phi$ lives on the real line, regardless of what the domain of $f$ is. 
Since the composition of $C^1$ functions is $C^1$ (chain rule), it remains to show that $\phi\in C^1$. Which amounts to direct computation of $\phi'(t)$ for $t\ne 0$, coupled with the observation that 
$$\phi'(0) = \lim_{t\to0}\frac{\phi(t)}{t}=1$$

For the sake of completeness, I prove a more general statement which also handles the vector-valued case. 
Claim. If $g:\mathbb R^n\to\mathbb R$ is continuous on $\mathbb R^n$ and $C^1$ on $\mathbb R^n\setminus\{0\}$, then the map  $\phi(x) = xg(x)$ is  $C^1$ on $\mathbb R^n$.
Proof. We can focus on one component of $\phi$, say $\phi_i(x) = x_i g(x)$. By the chain rule, 
$$\nabla \phi_i(x) = e_i g(x) + x_i \nabla g(x),\quad x\ne 0$$ where $e_i$ is the $i$th basis vector. Therefore, 
$\nabla \phi_i$ is continuous on $\mathbb R^n\setminus\{0\}$ and has limit $e_i g(0)$ as $x\to 0$. Also, 
observe that $$\phi_i (x) = x_i g(0) + o(|x|),\quad x\to 0$$
which implies $\phi_i$ is differentiable at $0$, with the derivative $e_i g(0)$. $\Box$
