The solutions are $f(x)=a-\frac{b} {x}$ and $g(x)=ax+\frac{b} {x}+c$ for some constants $a, b, c$. Let $P(x, y)$ denote the given assertion. Observe that we can find reals $a, b, c$, such that $g(1)=a+b+c$, $g(2)=2a+\frac{b} {2}+c$, $g(4)=4a+\frac{b} {4}+c$. Then we can consider $f_1(x)=f(x)-a+\frac{b} {x}$ and $g_1(x)=g(x)-ax-\frac{b} {x}-c$, which are easily checked to work and satisfy $g_1(1)=g_1(2)=g_1(4)=0$. Hence, we can assume WLOG that $g(1)=g(2)=g(4)=0$ and we will show that $f, g \equiv 0$, which will imply the desired solution set. Observe that $P(x, 1)$, $P(x, 2)$, $P(x, 4)$ give us $$g(x)=(x-1)f(x)=(x-2)f(2x)=(x-4)f(4x).$$ On the other hand, $P(2x, 1)$ and $P(2x, 2)$ give us $$g(2x)=(2x-1)f(2x)=(2x-2)f(4x).$$ Fix some $x \neq \frac{1}{2}, 1, 2, 4$ and assume that $f(x) \neq 0$. This means that $f(x), f(2x), f(4x) \neq 0$, so $$\frac{f(2x)}{f(4x)}=\frac{2x-2}{2x-1}=\frac{x-4}{x-2},$$ i.e. $x=0$, contradiction. Thus, $f(x)=g(x)=0$ for all $x \neq \frac{1}{2}, 1, 2, 4$. This easily implies that $f, g \equiv 0$: for $k \in \{\frac{1}{2}, 1, 2, 4\}$, $P(3, \frac{k} {3})$ implies $f(k)=0$ and $P(1, \frac{1}{2})$ implies $g(\frac{1}{2})=0$, so we are done.