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 \[\alpha^2 + \beta^2 = \left( \frac{1 + \sqrt{5}}{2} \right)^2 + \left( \frac{1 - \sqrt{5}}{2} \right)^2 = \boxed{3}.\]