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. $f(x) \in\left\{0, x^{2}-\frac{1}{x}\right\}$ for all $x \in \mathbb{R} \backslash\{0\}$. 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 $f(x)^{2}=x\left(x+\frac{1}{x}+1\right)\left(f(x)+f\left(\frac{1}{x}\right)\right)$ and $-f(x) f\left(\frac{1}{x}\right)=\left(x+\frac{1}{x}+1\right)\left(f(x)+f\left(\frac{1}{x}\right)\right)$. 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 $f(x)^{2}=x\left(x+\frac{1}{t x}+t\right)\left(f(x)+f\left(\frac{1}{t x}\right)\right)$ and $-f(x) f\left(\frac{1}{t x}\right)=t\left(x+\frac{1}{t x}+t\right)\left(f(x)+f\left(\frac{1}{t x}\right)\right)$. 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 ladder 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.