jmc

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 $$\begin{array}{rl}{x}^{2}f(x)+f(1-x)& =-x^4+2x,\\ (1-x{)}^{2}f(1-x)+f(x)& =-x^4+4x^3-6x^2+2x+1.\end{array}$$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 $x^2-x-1\ne 0,$ then $$f(x)=-(x^2-1)=1-x^2.$$Thus, if $x^2-x-1\ne 0,$ 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 $x=b$: $$\begin{array}{rl}\frac{3+\sqrt{5}}{2}f(a)+f(b)& =\frac{-5-\sqrt{5}}{2},\\ \frac{3-\sqrt{5}}{2}f(b)+f(a)& =\frac{-5+\sqrt{5}}{2}.\end{array}$$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 $$\begin{array}{rl}\frac{3+\sqrt{5}}{2}f(b)+f(a)& =\frac{3-\sqrt{5}}{2}\left(\frac{-5-\sqrt{5}}{2}-\frac{3+\sqrt{5}}{2}a\right)+f(a)\\ & =\frac{-5+\sqrt{5}}{2}.\end{array}$$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 $${\alpha }^2+{\beta }^2={\left(\frac{1+\sqrt{5}}{2}\right)}^2+{\left(\frac{1-\sqrt{5}}{2}\right)}^2=\boxed{3}.$$