Find all $f(x)$ such that $x(f(x+1)-f(x))=f(x)$ The problem
Find all $f(x): \mathbb{R} \to \mathbb{R} $ such that $x(f(x+1)-f(x))=f(x)$ and $|f(x) -f(y)| \le |x-y|, \forall x,y \in \mathbb{R}$
My approach
Obviously $f(x)=x$ is one solution, I suspect it is the only one. Now I substitute $x=0$ in the first condition, I get $f(0)=0$.
Thus substitute $y=0$ in the second condition I get $|f(x) | \le | x |  \tag{1}$
Another thing I got from the first condition:
\begin{align}
\frac{f(x+1)}{x+1}=\frac{f(x)}{x}
\end{align}
So I denote $c_x= x -\lfloor{x}\rfloor$, and assign a random function $f(x)=k_xx, \forall x\in [0,1], k_x$ a constant chosen independently for $c_x$, then $f(x)=k_xx, \forall x\in\mathbb{R}$.
Now what I can conclude is that $k_x$ all are smaller than $1$, because of $(1)$. But how do I proceed? Should I plug it into the second condition of the problem and try to deduce something?
Any help is appreciated!
EDITS
Earlier I mistakenly assumed that $f(x)$ is a polynomial. It's not. It is a function
 A: I will show that the only such functions satisfying the conditions in your question are:
$$
f(x) = ax, \;\text{where $-1 \leq a \leq 1$}
$$

Let $f$ be a $1$-Lipschitz function satisfying $x(f(x+1) - f(x)) = f(x)$. Let $f(1) = a$.
Claim. For any $x \in [0,1]$, we have $f(x) = ax$.
Proof. Suppose $f(c) = b$ for some $c \in (0,1]$. We observe that:
$$
f(c + 1) = \frac{c + 1}{c}f(c) = \frac{c+1}{c}b \\
f(c + 2) = \frac{c + 2}{c + 1}f(c + 1) = \frac{c+2}{c}b \\
f(c + 3) = \frac{c + 3}{c + 2}f(c + 2) = \frac{c+3}{c}b \\
\vdots
$$
Thus, one can prove by induction that $f(c + n) = \frac{c + n}{c}b$ (in particular, $f(n) = na$). Now if $c \in (0,1)$ and $b \neq ac$, then we have:
\begin{align*}
|f(c + n) - f(n)| &= \left|\frac{c + n}{c}b - an\right| \\
&= \left|b + \frac{n}{c}(b - ac)\right|
\end{align*}
Thus, as $n$ gets arbitrarily large, $|f(c + n) - f(n)|$ gets arbitrarily large, yet $|(c + n) - n| = c$ remains constant, which contradicts the $1$-Lipschitzness of $f$. $\blacksquare$
For any $x > 0$, we may write $x = n + c$ for some $n \in \mathbb{Z}_{\geq 0}$ and $c \in [0,1)$. If $c = 0$, then as mentioned above we have $f(n) = an$. If $c > 0$, then:
$$
f(x) = f(n + c) = \frac{n + c}{c}f(c) = \frac{n + c}{c}ac = a(n + c) = ax
$$
Therefore, $f(x) = ax$ for $x \geq 0$. The case for $x < 0$ can be done similarly: Observe that we have:
$$
\frac{f(x - 1)}{x - 1} = \frac{f(x)}{x}
$$
Therefore, we may again prove by induction that for $c \in (0,1]$:
$$
f(c - n) = \frac{c - n}{c}f(c)
$$
Any $x < 0$ may be written as $x = c - n$ for some $c \in (0,1)$ and $n \in \mathbb{Z}^+$, so:
$$
f(x) = f(c - n) = \frac{c - n}{c}f(c) = \frac{c - n}{c}(ac) = a(c - n) = ax
$$
so $f(x) = ax$ for $x < 0$. Finally, to see that we must have $-1 \leq a \leq 1$, we simply observe that:
$$
|a| = |a||1 - 0| = |a - 0| = |f(1) - f(0)| \leq |1 - 0| = 1
$$
by the $1$-Lipschitzness of $f$.
A: The inequality $|f(x) - f(y)| \le |x-y|$ implies $|f'(x)| \le 1$ for all $x \in \mathbb{R}$. Since the only polynomials which satisfy that take the form $f(x) = a + bx$, plugging this back into $x(f(x+1）- f(x)) = f(x)$ only reveals solutions $(a, b) = (0, b)$ for free $b$, or $f(x) = bx$.
A: Equation is $xf(x+1)=(x+1)f(x)$
And so $f(x+n)=(x+n)h(x)$ $\forall x\notin\mathbb Z$ and $\forall n\in\mathbb Z$ for some function $h(x)$
So $f(x+n)-f(y+n)=xh(x)-yh(y)+n(h(x)-h(y))$ and second inequation implies $h(x)$ constant over $\mathbb R\setminus\mathbb Z$
So $f(x)=cx$ $\forall x\notin \mathbb Z$ and (second line implies continuity), $f(x)=cx$ $\forall x$
hence the answer $\boxed{f(x)=cx\quad\forall x}$ which indeed is a solution, whatever is $c\in[-1,+1]$
