Solve for $f(x)$ if $f(f(x))=6x-f(x)$

For any $x$ define a sequence, $$a_0=x,a_1=f(x),a_2=f(a_1) \cdots, a_{n+1}=f(a_n) $$ Putting, $a_{n}$ in equation gives, $$ a_{n+2}=6a_n-a_{n+1}$$ This is a sequence, using characteristic equation method to solve this sequence. Since $f(x)>0$ gives, $$ a_{n} = 2^nx$$

This give $a_1=2x$ Checking this in functional equation, we see that it satisfies. And so $f(x)=2x$ is only function which satisfies the FE $\Box$

This approach is fatally flawed.

Here's an alternate solution via linear algebra. Let $V=\mathbb{R}\to\mathbb{R}$ denote the vector space of real functions on the real line and define $T\in V\to V$ to be the linear transformation $$(Tg)(x)=g(f(x))$$ Finally, let $p\in V$ denote the identity function, i.e. $p(x)=x$. Then your functional equation states $$T^2p+Tp=6p$$ Rearranging, $$0=(T^2+T-6)p=(T+3)(T-2)p$$ From the general theory of linear transformations, we know that $p$ is a sum of eigenfunctions: $p(x)=q(x)+r(x)$, where \begin{gather*} 0=(T+3)q(x)=q(f(x))+3q(x) \\ 0=(T-2)r(x)=r(f(x))-2r(x) \end{gather*}

If we are very lucky, $p$ is in fact an eigenfunction itself: $$0=(T+3)p=f(x)+3x$$ or $$0=(T-2)p=f(x)-2x$$ We can check that each of these are solutions to conditions (1-2) and only one solves condition (3). But I do not see how to rule out other solutions arising from linear combinations; for examples where this occurs, see the comments.