Prove $(y-x^2)$ is a prime ideal in $\mathbb{R}[x,y]$, but not maximal.

Indeed, $x\mapsto x$, $y\mapsto x^2$ gives us a homomorphism $\Bbb R[x,y]\to\Bbb R[x]$. Show that the kernel is $(x^2-y)$ and you are done as that shows $\Bbb R[x,y]/(x^2-y)\cong \Bbb R[x]$, which is an integral domain, but not a field.

Yes, the quotient is $\Bbb R[x]$. One way to see this is to consider the homomorphism $\phi:\Bbb R[x,y]\to\Bbb R[x]$ with $\phi(f(x,y))=f(x,x^2)$. This homomorphism is surjective, and its kernel is the ideal $(y-x^2)$, so by the First Isomorphism Theorem for rings, $\Bbb R[x,y]/(y-x^2)\cong \Bbb R[x]$.

As $\Bbb R[x]$ is not a field, then $(y-x^2)$ is maximal. But ine can write down lots of maximal ideals containing this ideal, for example $(x-a,y-a^2)$ for any $a\in\Bbb R$.