Is $\{f \in \mathbb{R}[x,y] \mid f(1,0) = f(0,1) = 0\}$ a prime ideal?
I don't believe so, but I don't quite trust my reasoning.
For $I = \{f \in \mathbb{R}[x,y] \mid f(1,0) = f(0,1) = 0 \}$ to be a prime ideal, we require that either $\mathbb{R}[x,y]/I$ to be an integral domain or that, for all $ab \in I$, either $a \in I$ or $b \in I$. However, if $ab \in I$, then this implies that $(ab)(1,0) = (ab)(0,1) = 0$. It could then be the case that the polynomial $a$ at $(1,0)$ is $0$ but is nonzero at $(0,1)$, and likewise the polynomial $b$ could be $0$ at $(0,1)$ but nonzero at $(1,0)$. Therefore, we have that $ab \in I$ but neither $a$ nor $b$ in I.
Your idea is correct. Consider for example $f(x,y) = x - 1$ and $g(x,y) = y - 1$.