<s>[INST] Functional equation with cyclic function. Find all functions $f:\mathbb R \to \mathbb R$ that satisfy:
$$ f(x)  + 3 f\left( \frac {x-1}{x} \right) = 7x.$$

Some progress: I plugged-in $\dfrac{x-1}{x}$ and $\dfrac{1}{1-x}$, got a system of equations and solving I got $f(x) = \dfrac{x^3-4x^2-3x-3}{4x^2-4x}$. But after testing it back looks like it's not sufficient.
 [/INST] Set $T(x) = \frac{x}{x-1}$. Then the composition $T^3(x) = x$. We have $f(Tx) = a x + b f(x)$ for $ a= \frac{7}{3}$ and $b = -\frac{1}{3}$, so
\begin{align*}
f(x)
&= f(T^3(x)) \\
&= a T^2(x) + b f(T^2 x) \\
&= a T^2(x) + a b T(x) + b^2 f(Tx) \\
&= a T^2(x) + a b T(x) +a b^2 x + b^3 f(x).
\end{align*}
That relation gives $f$ as a rational function of $x$.
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