First, let $f(f(x)f(1-x))=f(x)$ be eq. (1) and $f(f(x))=1-f(x)$ be eq. (2). Now, observe eq. (2). Notice that \[f(f(x))=1-f(x)\]\[\implies f(f(f(x)))=f(1-f(x))\]\[\iff 1-f(f(x))=f(1-f(x))\](by subtituting $x \to f(x)$ for eq. (2)) \[\iff f(x)=f(1-f(x)) \ \ ... \ (*)\](based on eq. (2)). Now, observe eq. (1). Notice that $f(f(x)f(1-x))=f(x)$ and $f(f(1-x)f(1-(1-x)))=f(1-x) \iff f(f(1-x)f(x))=f(1-x)$ (by subtituting $x \to 1-x$ for eq. (1)), so we get $f(x)=f(1-x) \ \ ... \ (**)$. After that, we get \[f(x) = f(1-x)\]\[\implies f(f(x))=f(1-f(x))\](by subtituting $x \to f(x)$ for eq. $(**)$) \[\iff f(f(x))=f(x)\](based on eq. $(*)$), which implies $f(f(x))=1-f(x) \iff f(x)=1-f(x) \iff f(x)=\frac{1}{2}$ (based on eq. (2)). Thus, $f(x)=\frac{1}{2}$ for all real number $x$ and it's easy to see that this function satisfies all of these equations.

Let's say $(1)=f(f(x)f(1-x))=f(x)$ and $(2)=f(f(x))=1-f(x)$. And let $Im(f)=\mathbb{S}$. From $(2)$ we get $f(x)=1-x$ for all $x\in \mathbb{S}$, so we get $1-x\in\mathbb{S}$ also, when $x\in\mathbb{S}$. In $(1)$ choose $x$ as element of $\mathbb{S}$, we get $f((1-x)\cdot x)=1-x$. Since $1-x$ is also element of $\mathbb{S}$, we get $f(x\cdot (1-x))=x$. So we get $x=1-x \implies x=\frac{1}{2}$. So we get $f(x)=\frac{1}{2}$ for all $x\in\mathbb{R}$.

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Let $P(x,y)$ and $Q(x)$ denote the first and second equations respectively. Easy to that any fixed point is equal to $.5.$ $P(x,1-x)\implies f(1-x)=f(x).$ $Q(f(x))\implies f(f(f(x)))=f(1-f(x))=f(f(x)).$ So $f(f(x))=.5,$ plugging in $Q(x)\implies \boxed{f(x)=.5},$ easy to check that it works.

The answer is $f(x)=\frac{1}{2}$ for all $x \in \mathbb{R}$. 1)$f(x)=f(1-x)$: By swapping $x$ with $1-x$ we easily get \[f(x)=f(f(x)f(1-x))=f(f(1-x)f(x))=f(1-x)\]2)$f(1-f(x))=1-f(x)$: From 1) we easily get \[f(f(x))=f(1-f(x))=1-f(x)\]3) The triple involution trick $f(f(f(x)))=f(x)$: Replacing $x$ with $f(x)$ we get \[f(f(f(x)))=1-f(f(x))=1-(1-f(x))=f(x)\]And finally combining all three we get \[f(x)=f(f(f(x)))=f(1-f(x))=1-f(x)\]$\Rightarrow$ \[2f(x)=1 \Rightarrow f(x)=\frac{1}{2}\]for all $x \in \mathbb{R}$ and we are done