So I already proved that $I$ is an ordinary ideal, now it's time to prove that it's not a prime ideal. So I need to show that $R/P$ is not an integral domain, but I'm confused as to how to go about doing that because by definition of $I$, its only elements I know of are the fact that it's $0$ at $f(1/2)$ and $f(1/3)$. To be an integral domain, $1\ne0$ and if $a,b$ are in $R/P$ and neither $a$ nor $b$ equals $0$, then $ab\ne0$.
I guess I'm confused about the structure of $R/P$? I know it's the cosets of $P$ in $R$, but that can't even be since every value of $R/P$ at $x=1/2$ and $x=1/3$ gets zeroed out. Then does it fail to be a ring, and therefore cannot be an integral domain?
Hint: consider two functions $f,g$ such that $f(1/2)=g(1/3)=0$ but neither $f(1/3)$ nor $g(1/2)$ is zero. Then $[f]$ and $[g]$ are nonzero in $R/I$ but their product is zero.