Find all functions $f:\mathbb{R}^+\to \mathbb{R}$ such that $xf(xf(x)-4)-1=4x$ 
Find all functions $f:\mathbb{R}^+\to \mathbb{R}$ such that for all $x\in\mathbb{R}^+$ the following is valid:
$$xf\big(xf(x)-4\big)-1=4x$$


All I could do is:

*

*$f(x)> {4\over x}$ for all $x$ so $f(x)>0$ for all $x$.

*$(4,\infty )\subseteq {\rm Range}(f)$, since $$f(xf(x)-4)={4x+1\over x} >4$$

*Function $g(x)=xf(x)-4$ is injective:
\begin{align}g(x_1)=g(x_2) &\implies  f(g(x_1))=f(g(x_2))\\&\implies {4x_1+1\over x_1}={4x_2+1\over x_2} \\&\implies x_1=x_2\end{align}

*Function $g$ satisfies $$\boxed{xg(g(x)) -(4x+1)g(x)+4x=0}$$
 A: Partial answer:
Consider the equation $xf(xf(x)-a)-1=ax$ for $a>0$ so that $$f(xf(x)-a)=a+\frac1x.$$ This means that $\lim\limits_{x\to+\infty}f(xf(x)-a)=a$ so that $\lim\limits_{x\to+\infty}f(x)=a$. Further, we have $$\lim_{x\to0^+}f(xf(x)-a)=+\infty$$ and since $f(x)>a/x\implies\lim\limits_{x\to0^+}f(x)=+\infty$, it follows that $\lim\limits_{x\to0^+}xf(x)=a$.
Let $m,n$ be integers such that $m<-1$ and $n>0$. Notice that $$f(x)=\sum\limits_{k=m}^na_kx^k$$ implies $\lim\limits_{x\to0^+}xf(x)=a$ so $a_{-1}=a$ and $a_i=0$ for all $m\le i<-1$. Likewise we have $\lim\limits_{x\to+\infty}f(x)=a$ so $a_0=a$ and $a_j=0$ for all $0<j\le n$. Thus if $f$ is a finite Laurent polynomial then the only solution to the functional equation is $$f(x)=a+\frac ax.$$
A: Partial answer
Let $g(x)=xf(x)-4$.
Our functional equation becomes

$$\frac{g\circ g(x)+4}{g(x)}=4+\frac1x$$

If there exists an invertible function $\phi(x):\mathbb R^+\to\mathbb R^+$ such that $g(x)=\phi(\phi^{-1}(x)+1)$, direct substitution gives
$$\frac{\phi(z+2)+4}{\phi(z+1)}=4+\frac1{\phi(z)}$$
or $$\phi(z+2)=\left[4+\frac1{\phi(z)}\right]\phi(z+1)-4$$
by substituting $z=\phi^{-1}(x)$.
Clearly, this recurrence relation extends a $\phi(x)$ defined arbitrarily on $(0,2)$ to the whole $\mathbb R^+$. By computing $\phi^{-1}(x)$ this method generates a large class of solution to the functional equation.
(Of course there are certain restrictions on $\phi(x)$ on $(0,2)$ so that $\phi(x)$ is invertible.)
The solution $f(x)=4+\frac4x$ corresponds to $g(x)=4x$ and $\phi(x)=k\cdot 4^x$ where $k$ is a positive constant.
A feature of this special solution is that $\phi(x)$ is continuous. In contrast, if rather arbitrary values are assigned to $\phi(x)$ on $(0,2)$, a discontinuous solution may yield. Below is the graph of $\phi(x)$ where $\phi(x) = x+1$ on $(0,2)$.

