, 2022

Solution:
The answer is either $f(x) = 0$ for all $x$ or $f(x) = x^2 - \frac{1}{x}$ for all $x$. These can be checked to work.

Now, I will prove that these are the only solutions. Let $P(x, y, z)$ be the assertion of the problem statement.

Lemma 1. $f(x) \in \{0, x^2 - \frac{1}{x}\}$ for all $x \in \mathbb{R} \setminus \{0\}$.

Proof. $P(1,1,1)$ yields $f(1) = 0$. Then, $P\left(x, 1, \frac{1}{x}\right)$ and $P\left(1, x, \frac{1}{x}\right)$ yield
$$\begin{aligned} f(x)^2 &= x\left(x + \frac{1}{x} + 1\right)\left(f(x) + f\left(\frac{1}{x}\right)\right) \\ -f(x) f\left(\frac{1}{x}\right) &= \left(x + \frac{1}{x} + 1\right)\left(f(x) + f\left(\frac{1}{x}\right)\right) \end{aligned}$$
Thus, we have $f(x)^2 = -x f(x) f\left(\frac{1}{x}\right)$, so we have $f(x) = 0$ or $f\left(\frac{1}{x}\right) = -\frac{f(x)}{x}$. Plugging in the latter into the first equation above gives us
$$f(x)^2 = x\left(x + \frac{1}{x} + 1\right)\left(f(x) - \frac{f(x)}{x}\right)$$
which gives us $f(x) = 0$ or $f(x) = x^2 - \frac{1}{x}$. This proves Lemma 1.

Lemma 2. If $f(t) = 0$ for some $t \neq 1$, then we have $f(x) = 0$ for all $x$.

Proof. $P\left(x, t, \frac{1}{t x}\right)$ and $P\left(t, x, \frac{1}{t x}\right)$ give us
$$\begin{aligned} f(x)^2 &= x\left(x + \frac{1}{t x} + t\right)\left(f(x) + f\left(\frac{1}{t x}\right)\right) \\ -f(x) f\left(\frac{1}{t x}\right) &= t\left(x + \frac{1}{t x} + t\right)\left(f(x) + f\left(\frac{1}{t x}\right)\right) \end{aligned}$$
Thus we have $t f(x)^2 = -x f(x) f\left(\frac{1}{t x}\right)$, so $f(x) = 0$ or $f\left(\frac{1}{t x}\right) = -\frac{t}{x} f(x)$. Plugging in the latter into the first equation gives us
$$f(x)^2 = x\left(x + \frac{1}{t x} + t\right)\left(f(x) - \frac{t f(x)}{x}\right)$$
which gives us either $f(x) = 0$ or $f(x) = x\left(x + t + \frac{1}{t x}\right)\left(1 - \frac{t}{x}\right) = x^2 - \frac{1}{x} - \left(t^2 - \frac{1}{t}\right)$. Note that since the latter expression doesn't equal $x^2 - \frac{1}{x}$, since $t \neq 1$, we must have that $f(x) = 0$. Thus, we have proved Lemma 2.

Combining these lemmas finishes the problem.

Solution:
Suppose $x y z = 1$ and $x + y + z \neq 0$ and that $x, y, z$ are not all the same. Then we have
$$\begin{aligned} & f(x)^2 - f(y) f(z) = x(x + y + z)(f(x) + f(y) + f(z)) \\ & f(y)^2 - f(z) f(x) = y(x + y + z)(f(x) + f(y) + f(z)) \\ & f(z)^2 - f(x) f(y) = z(x + y + z)(f(x) + f(y) + f(z)) \end{aligned}$$
Squaring the first equation and subtracting the second equation times the third gives us: $f(x) F(x, y, z) = (x^2 - y z) G(x, y, z)^2$, where $F(x, y, z) = f(x)^3 + f(y)^3 + f(z)^3 - 3 f(x) f(y) f(z)$ and $G(x, y, z) = (x + y + z)(f(x) + f(y) + f(z))$. If $F(x, y, z) = 0$, it is not too hard to see that we get $f(x) = f(y) = f(z) = 0$. If not, then we can let $K = \frac{G^2}{F}$ and we substitute $(f(x), f(y), f(z)) = (K(x^2 - y z), K(y^2 - x z), K(z^2 - x y))$ into the first equation to get $K^2 x(x^3 + y^3 + z^3 - 3 x y z) = K x(x^3 + y^3 + z^3 - 3 x y z)$. Thus, we have $K = 0$ or $K = 1$. Thus, we have either $(f(x), f(y), f(z)) = (0, 0, 0)$ or $(f(x), f(y), f(z)) = (x^2 - y z, y^2 - z x, z^2 - x y)$.

Thus, $f(0.5) = 0$ or $f(0.5) = 0.5^2 - \frac{1}{0.5}$. If the former is true, then for all $y$ and $z$ such that $y z = 2$ and $y + z \neq 0.5$, we have $f(y) = 0$. However, this gives that $f(y) = 0$ for all $y$. Likewise, if the latter were true, we would have $f(y) = y^2 - \frac{1}{y}$ for all $y$, so we are done.