The function $f : \mathbb{R} \to \mathbb{R}$ satisfies
\[x^2 f(x) + f(1 - x) = -x^4 + 2x\]for all real numbers $x.$  Then $f(x)$ can be uniquely determined for all values of $x,$ except $f(\alpha)$ and $f(\beta)$ for some real numbers $\alpha$ and $\beta.$  Compute $\alpha^2 + \beta^2.$
Replacing $x$ with $1 - x,$ we get
\[(1 - x)^2 f(1 - x) + f(x) = -(1 - x)^4 + 2(1 - x) = -x^4 + 4x^3 - 6x^2 + 2x + 1.\]Thus, $f(x)$ and $f(1 - x)$ satisfy
\begin{align*}
x^2 f(x) + f(1 - x) &= -x^4 + 2x, \\
(1 - x)^2 f(1 - x) + f(x) &= -x^4 + 4x^3 - 6x^2 + 2x + 1.
\end{align*}From the first equation,
\[x^2 (1 - x)^2 f(x) + (1 - x)^2 f(1 - x) = (1 - x)^2 (-x^4 + 2x) = -x^6 + 2x^5 - x^4 + 2x^3 - 4x^2 + 2x.\]Subtracting the second equation, we get
\[x^2 (1 - x)^2 f(x) - f(x) = -x^6 + 2x^5 - 2x^3 + 2x^2 - 1.\]Then
\[(x^2 (1 - x)^2 - 1) f(x) = -x^6 + 2x^5 - 2x^3 + 2x^2 - 1.\]By difference-of-squares,
\[(x(x - 1) + 1)(x(x - 1) - 1) f(x) = -x^6 + 2x^5 - 2x^3 + 2x^2 - 1,\]or
\[(x^2 - x + 1)(x^2 - x - 1) f(x) = -x^6 + 2x^5 - 2x^3 + 2x^2 - 1.\]We can check if $-x^6 + 2x^5 - 2x^3 + 2x^2 - 1$ is divisible by either $x^2 - x + 1$ or $x^2 - x - 1,$ and we find that it is divisible by both:
\[(x^2 - x + 1)(x^2 - x - 1) f(x) = -(x^2 - x + 1)(x^2 - x - 1)(x^2 - 1).\]Since $x^2 - x + 1 = 0$ has no real roots, we can safely divide both sides by $x^2 - x + 1,$ to obtain
\[(x^2 - x - 1) f(x) = -(x^2 - x - 1)(x^2 - 1).\]If $x^2 - x - 1 \neq 0,$ then
\[f(x) = -(x^2 - 1) = 1 - x^2.\]Thus, if $x^2 - x - 1 \neq 0,$ then $f(x)$ is uniquely determined.

Let $a = \frac{1 + \sqrt{5}}{2}$ and $b = \frac{1 - \sqrt{5}}{2},$ the roots of $x^2 - x - 1 = 0.$  Note that $a + b = 1.$  The only way that we can get information about $f(a)$ or $f(b)$ from the given functional equation is if we set $x = a$ or $x = b$:
\begin{align*}
\frac{3 + \sqrt{5}}{2} f(a) + f(b) &= \frac{-5 - \sqrt{5}}{2}, \\
\frac{3 - \sqrt{5}}{2} f(b) + f(a) &= \frac{-5 + \sqrt{5}}{2}.
\end{align*}Solving for $f(b)$ in the first equation, we find
\[f(b) = \frac{-5 - \sqrt{5}}{2} - \frac{3 + \sqrt{5}}{2} f(a).\]Substituting into the second equation, we get
\begin{align*}
\frac{3 + \sqrt{5}}{2} f(b) + f(a) &= \frac{3 - \sqrt{5}}{2} \left( \frac{-5 - \sqrt{5}}{2} - \frac{3 + \sqrt{5}}{2} a \right) + f(a) \\
&= \frac{-5 + \sqrt{5}}{2}.
\end{align*}This means that we can take $f(a)$ to be any value, and then we can set
\[f(b) = \frac{-5 - \sqrt{5}}{2} - \frac{3 + \sqrt{5}}{2} f(a)\]to satisfy the functional equation.

Thus, $\alpha$ and $\beta$ are equal to $a$ and $b$ in some order, and
\[\alpha^2 + \beta^2 = \left( \frac{1 + \sqrt{5}}{2} \right)^2 + \left( \frac{1 - \sqrt{5}}{2} \right)^2 = \boxed{3}.\]