= Solution
The <chain rule> along the specified straight line gives
$$
F'(0)=\alpha f_1(a,b)+\beta f_2(a,b),\qquad
F''(0)=\alpha^2f_{11}(a,b)+2\alpha\beta f_{12}(a,b)+\beta^2f_{22}(a,b).
$$
Here equality of mixed <partial derivatives> follows from smoothness. Apply <Taylor theorem> in the single variable $t$:
$$
F(t)=F(0)+tF'(0)+\tfrac12t^2F''(0)+o(t^2).
$$
Writing $h_1=x_1-a=\alpha t$ and $h_2=x_2-b=\beta t$ produces the linear and quadratic terms
$$
f(a+h_1,b+h_2)=f(a,b)+h_1f_1+h_2f_2
+\tfrac12(h_1^2f_{11}+2h_1h_2f_{12}+h_2^2f_{22})+o(|h|^2),
$$
with all displayed <partial derivatives> evaluated at $(a,b)$. For a remainder valid uniformly over directions, apply the <Taylor formula with integral remainder> to $s\mapsto f((a,b)+sh)$:
$$
f((a,b)+h)=f(a,b)+\nabla f(a,b)\cdot h
+\int_0^1(1-s)h^TH((a,b)+sh)h\,ds.
$$
Continuity of the <Hessian matrix> makes the difference between the integral and $\tfrac12h^TH(a,b)h$ be $o(|h|^2)$. The printed ellipsis is therefore justified as a finite <Taylor expansion> with a remainder. Smoothness alone does not assert convergence of an infinite <Taylor series>.
At a <stationary point>, the linear term vanishes. The symmetric <Hessian matrix> satisfies
$$
h^THh
=H_{11}\left(h_1+\frac{H_{12}}{H_{11}}h_2\right)^2
+\frac{\det H}{H_{11}}h_2^2.
$$
If $H_{11}>0$ and $\det H>0$, this is a <positive-definite quadratic form>. Its minimum over the <unit circle> is a positive number $c$, so $h^THh\geq c|h|^2$. The <Taylor remainder> is smaller than $c|h|^2/4$ for sufficiently small $h$, proving $f((a,b)+h)>f(a,b)$ for nonzero nearby $h$. Thus \b[the stationary point is a strict local minimum].
For the given polynomial, the <gradient> equations reduce to
$$
x_2=-x_1,\qquad 4x_1(x_1^2-1)=0.
$$
The <stationary points> are $(0,0),(1,-1),(-1,1)$, and
$$
H=\begin{pmatrix}12x_1^2-2&2\\2&2\end{pmatrix}.
$$
At both points with $x_1=\pm1$, we have $H_{11}=10>0$ and $\det H=16>0$. Consequently
$$
\boxed{(1,-1)\text{ and }(-1,1)\text{ are local minima, with }f=-1.}
$$
At the origin the <determinant> of the <Hessian matrix> is $-8$, giving a <saddle point>. An independent check is the identity $f=(x_1^2-1)^2+(x_1+x_2)^2-1$, which proves that the two displayed <local minima> are also the only global minima.
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