isomorphism $\psi$ between quotient rings 
Show that there is an isomorphism $\psi$ between the rings $R_1 := \mathbb{Q}[r,s]/(r+s-1, r-r^2, s-s^2)$ and $R_2 := \mathbb{Q}[y]/(y^2-1)$ so that $\psi([r]) = [(1+y)/2]$ and $\psi(s)=[(1-y)/2].$

I could define an surjective homomorphism $\psi$ from $\mathbb{Q}[r,s]$ to $\mathbb{Q}[y]/(y^2-1)$ and show that its kernel is $(r+s-1,r-r^2, s-s^2)$. The first isomorphism theorem for rings gives an isomorphism $\psi_2 : R_1\to R_2$ so that $\psi_2 \circ q = \psi,$ where $q : \mathbb{Q}[r,s]\to R_1$ is the quotient map. Then it suffices to define $\psi$ so that $\psi(r) = [(1+y)/2]$ and $\psi(s) = [(1-y)/2]$. Define $\psi$ by $\psi(f(r,s)) = [f(\frac{1+y}2, \frac{1-y}2)],$ where each $f(r,s)\in \mathbb{Q}[r,s].$ It's easy to show that $\psi$ is a homomorphism. I can also show that $(r+s-1, r-r^2, s-s^2)\subseteq \ker \psi$ and that $\psi$ is surjective using the fact that the elements of $\mathbb{Q}[y]/(y^2-1)$ are of the form $[ay+b]$ for some $a,b \in \mathbb{Q}$. This can be shown using the division algorithm for polynomials. Then one can find $f$ so that $f(\frac{1+y}2, \frac{1-y}2) =ax+b.$ But how do I show that $(r+s-1, r-r^2, s-s^2) = \ker \psi$?

I fear that your approach is more difficult than necessary. Observe that $\mathbb Q[y]$ is a principal ideal domain (PID), hence every ideal is generated by a single element. Consequently, if we can find a surjective ring homomorphism $\varphi : \mathbb Q[y] \to \mathbb Q[r, s]/(r + s - 1, r - r^2, s - s^2)$ such that $(y^2 - 1) \subseteq \ker \varphi,$ then the generator of $\ker \varphi$ must divide $y^2 - 1$ and must therefore be $y - 1,$ $y + 1,$ or $y^2 - 1.$ Based on your objective, you already have a candidate for $\varphi$: let $\varphi(f(y)) = f(2 \bar r - 1),$ where $\bar r$ denotes the image of $r$ in $R_1.$ Observe that $$\bar r = \frac{2 \bar r - 1 + 1} 2 = \varphi {\left(\frac{y + 1} 2 \right)} \text{ and } \bar s = 1 - \bar r = \varphi {\left(1 - \frac{y + 1} 2 \right)},$$ hence $\varphi$ is surjective. Further, we have that $\varphi(y^2 - 1) = (2 \bar r - 1)^2 - 1 = 4 \bar r^2 - 4 \bar r + 1 - 1 = 0,$ from which it follows that $(y^2 - 1) \subseteq \ker \varphi.$ Like I mentioned before, from here, we deduce that $\ker \varphi = (y - 1)$ or $\ker \varphi = (y + 1)$ or $\ker \varphi = (y^2 - 1).$ Once you verify that the former two cases cannot happen, you may conclude that $\ker \varphi = (y^2 - 1)$ so that $R_1 \cong R_2$ via $\varphi.$
Last, you must establish that $\varphi$ and $\psi$ are function inverses. Once you have done this, then you can conclude that $R_1 \cong R_2$ via $\psi,$ as desired.
