If $g\in C^k(\mathbb{R}^n,\mathbb{R})$ and $g(x) = f(x)(1+f(x)^4)$, then $f\in C^k(\mathbb{R}^n,\mathbb{R})$ I'm trying to solve this problem

Let $\ f,g:\mathbb{R}^n \rightarrow \mathbb{R}$ be two functions, and suppose $g$ is $C^k$ ($k\geq 1$). If $g(x) = f(x)(1+f(x)^4)$ for all $x\in\mathbb{R}^n$, then $f$ is also $C^k$.

I am trying to write $f$ in terms of $g$, but I had no succes in it. I thought that maybe the Implicit Function Theorem could be useful here, but I can't find out how to use it in the problem. 
Any tips?
 A: Let $h(x,y)=g(x)-y(1+y^{4})$, $x\in{\bf{R}}^{n}$, $y\in{\bf{R}}$, then $h$ is $C^{k}$ on the variable $y$, and $\dfrac{\partial f}{\partial y}(x,y)=-(1+5y^{4})\ne 0$, so Implicit Function Theorem says that, locally $U_{x}$, there exists uniquely a $C^{k}$ function $\widetilde{f}$ on $U_{x}$ such that $h(x,\widetilde{f}(x))=0$. Since $f$ also satisfies the equation, so $f=\widetilde{f}$ on $U_{x}$, so $f$ is $C^{k}$ on $U_{x}$. Since $U_{x}$ varies arbitrarily, so $f$ is $C^{k}$ throughout. 
A: Let $h(x)=x(1+x^4)$, it is not hard to see that $h$ is smooth, $h'(x) \ge 1$ and hence has a smooth inverse $h^{-1}$.
Since $g = h \circ f$ we see that $f = h^{-1} \circ g$ from which it follows that $f$ is as smooth as $g$.
A: $h(x)=x(1+x^4)$ is bijective, so $f$ is continuous if $g:x\mapsto h(f(x))$ is. 
Hence if $g$ is $C^1$ then $f$ is continuous, and
$\quad g'(x) = \left(5 f(x)^4+1\right) f'(x)$
the first factor of which is continuous, so $f'$ is continuous if $g'$ is.
If $g$ is $C^2$ then $f,f'$ are continuous, and by product rule
$\quad g''(x) = \left(5 f(x)^4+1\right) f''(x) +  \text{"finitely many continuous terms"}$
so $f''$ is continuous if $g''$ is.
Etcetera...
