Let $f:\mathbb{R}^n\to \mathbb{R}$ a fuction of class $C^1$ and $F:\mathbb{R}^n\times \mathbb{R}\to \mathbb{R}$ defined by, $$F(x,y)=y(1+y^6)-f(x).$$ Show that there is an unique fuction $g:\mathbb{R}^n\to \mathbb{R}$ of class $C^1$ such that, $$f(x)=g(x)(1+(g(x)^6).$$
My attempt:
I'd proved that for every $x\in \mathbb{R}^n$ there is an unique $y\in \mathbb{R}$ such that $F(x,y)=0$. Then, I apply the implicit function theorem to $F$ and for all $x\in \mathbb{R}^n$ I find a neighborhood $B_x$ of $x$ and an unique continuous fuction $g_x:B_x\to \mathbb{R}^n$ such that $g_x(x)=y$ and $$F(x,g_x(x))=0$$ that is, $$f(x)=g_x(x)(1+(g_x(x))^6.$$ How can I define a function $g:\mathbb{R}^n\to \mathbb{R}$ which satisface the conditions?
The function $\ \psi:\mathbb R \rightarrow \mathbb R\ $ defined by $\ \psi\left(y\right) = y \left(1 + y^6\right)\ $ for $\ y\in \mathbb R\ $ is strictly increasing (since $\ \psi'\left(y\right) = 1+7 y^6 > 0\ $ for all $\ y\in\mathbb R\ $), and $\ \psi\left(y\right) \rightarrow \infty, \psi\left(-y\right) \rightarrow -\infty, $ as $\ y\rightarrow\infty\ $. Therefore, for any $\ \xi $ in $\ \mathbb R\ $ there is a unique real number, $\ \gamma\left(\xi\right) $, such that $\psi\left(\gamma\left(\xi\right)\right) = \xi\ $ $\left(\ = \ \gamma\left(\xi\right) \left(1 + \gamma\left(\xi\right)^6\right)\right) \ $. It follows that $\ g\left(x\right)\ $ must be $\ \gamma\left(f\left(x\right)\right)\ $.