Interesting functional equations problem? $f(x) + 3 f\left( \frac {x-1}{x} \right) = 7x$

I think this is essentially the solution you came up with, but it's helpful to write out all the details.

Let $y=\frac{x-1}{x}$, $z=\frac{1}{1-x}$. Then, if $g(x)=\frac{x-1}{x}$, we have $y=g(x)$, $z=g(y)$, and $x=g(z)$. So the functional equation implies that \begin{align} f(x) + 3f(y)&=7x\\ f(y) + 3f(z)&=7y\\ f(z) + 3f(x)&=7z \end{align}

Then $f(y)=7y-3f(z)$ and $f(z)=7z-3f(x)$, so $f(y)=7y-3(7z-3f(x))=7y-21z+9f(x)$.

So $f(x)+21y-63z+27f(x)=7x$, meaning that \begin{align} f(x)&=\frac{7x-21y+63z}{28} \\ &=\frac{x-3y+9z}{4}\\ &=\frac{1}{4}\left(x-3+\frac{3}{x}+\frac{9}{1-x}\right) \end{align}

It's not too hard to check that this satisfies the original functional equation for all $x \neq 0,1$. (Actually, it's easier if you leave everything in terms of $x$, $y$, and $z$.)

What about the value of $f$ at $0$ and $1$? By setting $x=1$ in the original equation, we can see that:

$$ f(1)+3f(0)=7 \tag{1} $$

Because $\frac{x-1}{x}$ is never equal to $1$ and undefined at $x=0$, this is the only meaningful data the equation gives us about $f(0)$ and $f(1)$, so you can pick any values for them you wanted that satisfy $(1)$. There doesn't seem to be any choice that's much nicer than the others, since $f$ has simple poles at both $0$ and $1$ (e.g., we can never make $f$ continuous).