Put $y=0$ to get either $f$ constant, which is absurd, or $f(0)=0$. Put $x=y=\sqrt t$ for $t\ge0$ to get $f(t)=t$ for $t\ge0$. Now let $c=f(-1)$. Put $(x,y)=(t,-1)$ and $(x,y)=(-1,t)$ for $t<0$ to conclude $f(-t)=-t+|t+ct| = (|1+c|-1)t$. Let $k=1-|1+c|$. So we just need to find all $k$ such that the function $f$ defined by $f(x)=x$ for $x \ge 0$ and $f(x) = kx$ for $x \le 0$ works. Clearly it suffices to check when $xy<0$. By cases we see that this forces $|k^2-k|=|k-1|$ hence $k=\pm1$ both of which work. Hence the answers are the identity function and the absolute value function.

f(x) equals x or modx .

$\text{Claim 1:}$ $f(0)=0$ Proof:if $f(0)\neq 0$ $P(\frac{x}{f(0)},0)$ $\implies$ $f$ is constant. So Claim is proved. $\text{Claim 2:}$ $f(x)=x$ for all positive $x$ Proof: $f(x,x)$ $\implies$ $f(x^2)=x^2$ so for positive $x$ we have $f(x)=x$ $\text{Claim 3:}$ $f(x)=0$ $\iff$ $x=0$ Proof: Since $f(0)=0$ there exists $a$ s.t $f(a)=0$. $P(a,1)$ $\implies$ $a=0$ Now let's select positive $x$ and substitute $P(-x,-1)$ $\implies$ $f(f(-x)-xf(-1))=0$ so $f(-x)=xf(-1)$ So solutions are $f(x)=x$ for positives $f(x)=-xf(-1)$ for negatives.

Murad.Aghazade wrote: So solutions are $f(x)=x$ for positives $f(x)=-xf(-1)$ for negatives. Unless I'm mistaken, not all these functions work?

v_Enhance wrote: Murad.Aghazade wrote: So solutions are $f(x)=x$ for positives $f(x)=-xf(-1)$ for negatives. Unless I'm mistaken, not all these functions work? I think it is true whatever $f(-1)$ is. Edited:Actually we can find the value of $f(-1)$ looking at $P(5,-3)$ and $P(-3,5)$

$f(-x)=xf(-1)$ then set x=-a (a is positive) then $f(-1)=-1$

oh sorry you set x as positive

Can someone explain me the casework in post 2? I don't understand the part where he got $|k^2-1|=|k-1|$.

Let $P(x,y)$ be the assertion $f(xf(y)-yf(x))=f(xy)-xy$. $P(x,0)\Rightarrow f(xf(0))=f(0)$, so either $f$ is constant (no such solutions) or $f(0)=0$. $P(x,x)\Rightarrow f\left(x^2\right)=x^2$, so $f$ fixes the positive reals. From now on, let $x,y\in\mathbb R^+$. $P(-x,-y)\Rightarrow f(yf(-x)-xf(-y))=f(xy)-xy=0$ $P(yf(-x)-xf(-y),1)\Rightarrow yf(-x)-xf(-y)=0\Rightarrow\frac y{f(-y)}=\frac x{f(-x)}=c$ for some constant $c$. Thus $f(-x)=cx$ for some constant $c$. Testing this in the original equation, we find $c\in\{-1,1\}$, hence the solutions $\boxed{f(x)=x}$ and $\boxed{f(x)=|x|}$ which both work.

Nice problem! Plug in $(x,0)$ to get $f(xf(0))=f(0)$. So for all $r \in \mathbb{R}$, $f(\frac{r}{f(0)}*f(0))=f(r)=f(0)$, if $f(0) \neq 0$, meaning that function $f$ would be constant. However, plugging in any $x,y \neq 0$ would give $0 = -xy$, a contradiction. Therefore, we conclude that $f(0)=0$. So by now we have tried every pair including $0$. Now let's plug in $(x,x)$ to get $f(x^{2})=x^{2}$. Hence, for all nonnegative $t$, $f(t)=t$. Ok, seems like we're almost there. How can we take advantage of this? Well, to get a positive number you can either multiply two positive numbers or two negative numbers. Let's try testing two negative numbers $(-x,-y)$, while keeping $x,y \ge 0$ (if we test (x,y) we get $f(0)=0$ which is already found). We get $f(-xf(-y)+yf(-x))=0$. But what happens if $-xf(-y)+yf(-x) < 0$?, we wouldn't know the value of $f(-xf(-y)+yf(-x))$ then! Well be can plug in $(-y,-x)$ for the same values of $x,y$ to get $f(-yf(-x)+xf(-y))=0$. Notice that $-yf(-x)+xf(-y) = -(-xf(-y)+yf(-x))$, so at least one of them must be nonnegative. But we also know that $f(t)=t$ for all nonnegative integers $t$! Therefore one of $(-xf(-y)+yf(-x),-yf(-x)+xf(-y))$ must be equal to $0$. WLOG let $-xf(-y)+yf(-x)=0$. Then $xf(-y)=yf(-x)$ (recall that $x,y$ are both nonnegative). Well, what does this tell us? We have simply "fix" a value, say $f(-1)$ and put $f(-r)$ in terms of $f(-1)$ for all nonnegative $r$. Plugging in $(x,1)$, we get $f(-x)=f(-1)x$ for some constant $f(-1)$. Hence, function $f$ is linear. Well, how do I figure out the value of $f(-1)$? Plug in $(1,-1)$ and $(-1,1)$ to get $f(f(-1)+1)=f(-1)+1=f(-1-f(-1))$. Once again, we see that one of $(f(-1)+1,-1-f(-1))$ must be nonnegative. WLOG if $f(-1)+1$ were to be positive, then $-1-f(-1)$ is negative, but $f(f(-1)+1)=f(-1)+1=f(-1-f(-1))$, so $f(-x)=x$. WLOG if $f(-1)+1$ were to be 0, then $f(-1)=-1$, so $f(-x)=-x$. Hence, functions $f$ can either be $y=x$ or $y=|x|$. It is trivial to show that they satisfy the conditions, so we are done. $\blacksquare{}$

The only solutions are $\boxed{f(x) = x}$ and $\boxed{f(x) = |x|}$, which work. Let $P(x,y)$ denote the given assertion. $P(x,0): f(xf(0)) = f(0)$. If $f(0) \ne 0$, then $f$ is constant, absurd. So $f(0) = 0$. $P(x,x): f(x^2) = x^2$, so $f(x) = x\forall x\ge 0$. $P(x,-1): f(xf(-1) + f(x)) = f(-x) + x$. Claim: If $f(x) = 0$, then $x = 0$. Proof: Suppose there existed $f(a) = 0$ for some $a\ne 0$. $P(a,1): f(a) = -a $, so $a = 0$, contradiction. $\square$ If $x \le 0$, then $P(x,-1)\implies f(x f(-1) + f(x)) = 0$, so $f(x) = -xf(-1)$ for all $x\le 0$. $P(x, 1): f(x - f(x)) = f(x) - x$. For any $x < 0$, then $f(x + xf(-1)) = -x - xf(-1)$. Since $f(k) = -k\implies k \le 0$, we have that $x(f(-1) + 1) \le 0$, so $f(-1) + 1 \ge 0$. If $f(-1) + 1 > 0$, then $x + x f(-1)$ for $x\le 0$ can take any real number less than or equal to $0$, so $f(x) = -x = |x|\forall x\le 0$. If $f(-1) + 1 = 0$, then $f(-1) = -1$, so $f(x) = -x \cdot -1 = x$ for all $x\le 0$. This gives the desired solutions.