There is no such function. For $x=0$ we have $f(f(0)) = 1$; denote $f(0) = a$, so $f(a) = 1$. If $a=1$, then for $x=1$ we would have $1 = f(f(1)) + 1\cdot f(1) = 1+1=2$, absurd, so $a\neq 1$. Assume there exists $b$ with $f(b) = 0$; then for $x=b$ we would have $1 = f(f(b)) + b f(b) = f(0) + 0 = a$, absurd. Now, assume $f(x)=f(y)=c$; it follows $c(x-y)=0$, and since $c\neq 0$ it follows $x=y$, therefore $f$ is injective. Finally, for $x=a$ we have $1 = f(f(a)) + af(a) = f(1) + a$, so $f(1) = 1-a$, and for $x=1$ we have $1 = f(f(1)) + f(1) = f(1-a) + 1 - a$, whence $f(1-a) = a = f(0)$; from injectivity follows $1-a = 0$, i.e. $a=1$, forbidden.

There is no such function. For $x=0$ we have $f(f(0)) = 1$; denote $f(0) = t$, so $f(t) = 1$. For $x=t$ $f(f(t))+tf(t)=1$ which is $f(1)+t=1$ For $x=1$ $f(f(1))+f(1)=1$ . Therefore we get $f(f(1))=t=f(0)$ Suppose $f(x)=f(y)$ then either $f(x)$ is identically zero (not possible) or $f(x)$ is injective. So $f(1)=0$ . So $t=1$. So $1=f(t)=f(1)=0$ which is absurd . No function exists.

Let $P(x)$ be $f(f(x))+xf(x)=1$ $f(x)\neq 0(\forall x\in \mathbb R):$ If $\exists \alpha\text{ s.t. }f(\alpha)=0$ $P(\alpha)\implies f(0)=1$ $P(0)\implies f(1)=1$ $P(1)\implies f(1)=\frac{1}{2}$ This is absurd.Therefore $f(x)\neq 0(\forall x\in \mathbb R)$ $f$ is injective: If $f(a)=f(b)\neq 0,$considering $P(a),P(b),$then $af(a)=bf(b)\implies a=b$.Therefore $f$ is injective. Let $f(0)=t,$ $P(0)\implies f(t)=1$ $P(t)\implies f(1)=1-t$ $P(1)\implies f(1-t)=t$ Since $f$ is injective,$1-t=0\implies t=1$.Then $f(1)=0$ which is absurd. Therefore there is no such function.

Posting for fun, although ig it is basically the same as the above ones. Let $P(x)$ be the assertion. $P(0)\implies f(f(0))=1$ $P(f(0))\implies f(0)+f(1)=1$ $P(1)\implies f(f(1))+f(1)=1$, hence $f(f(1))=f(0)$. If $f(a)=f(b)$, then $f(a)(a-b)=0$, thus $a=b$ or $f(a)=0$. Therefore, $f(1)=0$ or $f(0)=0$. If $f(0)=1$, then $f(1)=f(f(0))=1$ and thus $1+1=1$, contradiction. If $f(0)=0$, then $0=f(0)=f(f(0))=1$, contradiction. Answer. No solutions.

x=0 ---> f(f(0))=1 x=f(0) ---> f(1)+f(0)=1 If f has a root r, x=r ---> f(0)=1 ---> 1=f(f(0))=f(1). But, f(1)=1-f(0)=0. Contradiction. So, f(x) is never 0 We have x=[1-f(f(x))]/f(x), which implies f is bijective. x=1 ---> f(f(1))=1-f(1)=f(0) ---> f(1)=0. But, 1=f(f(0))=f(1-f(1))=f(1) ---> 1=0. So, no such functions exist.

$x=0\implies f(f(0))=1$ $x=1\implies f(f(1))+f(1)=1$ $x=f(0)\implies f(0)+f(1)=1$ $\therefore f(0)=f(f(1))$ $x=f(1)\implies f(1)f(0)=0$ $f(0)=0\implies 1=f(f(0))=0$ $f(1)=0\implies 1=f(0)=f(f(0))=f(1)=0$ $\therefore \not \exists f$

$x=0: \enspace f(f(0)) = 1$ $x=f(0): \enspace f(1) + f(0) = 1$ $x=f(1)+f(0): \enspace 2f(1) + f(0) = 1$ Thus, $f(1) = 0$ and $f(0) = 1$. But then, $1 = f(f(0)) = f(1) = 0$, which is absurd. Hence, no such function exists.