$4f(x)+1=f(f(f(x)))=f(4x+1)$ . Set $x=-1/3$,then $4f(-1/3)+1=f(4\cdot (-1/3)+1)\Leftrightarrow f(-1/3)=-1/3$ It remains to show that $f(x_0)=x_0\Rightarrow x_0=-1/3$ Indeed $4x_0+1=f(f(x_0))=f(x_0)=x_0\Leftrightarrow x_0=-1/3$ qed

Solved with jelena_ivanchic and a2048 Note that if $f(a)=a$ and $f(b)=b$ then $a=b=-\frac{1}{3}$ so if $x$ exists, then it will be unique. To show such an $x$ exists, note that $f(f(x))=4x+1\implies f(4x+1)=4f(x)+1$. Now put $x=-1/3$ to get $f(-1/3)=-1/3$ so done.

We have $f(f(x))=4x+1$, so $f(x)-x$ is injective. Also, $f(f(f(x))=4f(x)+1=f(4x+1)$, so setting $x=-\frac13$, we get $f\left(-\frac13\right)=-\frac13$, and by the injectivity of $f(x)-x$, this is the unique fixed point.