$P(0,0): f(0)=0$ or $f(0)=2$ If $f(0)=0$, then $P(x,0): \boxed{f(x)=0}$ If $f(0)=2$: $P(\frac{2}{3},-\frac{2}{3}): f(\frac{2}{3})=0$ or $f(-\frac{2}{3})=0$ Thus, there exists $\alpha \neq 0$ such that $f(\alpha)=0$. $P(x-\alpha,\alpha): f(x)=-\frac{9}{2} \alpha (x-\alpha) \implies f$ is linear. Checking, we see $\boxed{f(x)=2+3x}$ or $\boxed{f(x)=2-3x}$.