= Solution
The <Hessian matrix> has entries
$$
f_{xx}=4(s-1)+8x^2,\qquad f_{xy}=8xy,\qquad f_{yy}=4(s-1)+8y^2+\frac{7(3-y^2)}{(3+y^2)^2}.
$$
At each <stationary point> the mixed entry vanishes, so the signs of the diagonal entries directly classify the local quadratic change:
$$
\begin{array}{c|c|c}
\text{point}&H&\text{nature}\\\hline
(\pm1,0)&\operatorname{diag}(8,7/3)&\text{strict local minima}\\
(0,0)&\operatorname{diag}(-4,-5/3)&\text{strict local maximum}\\
(0,\pm1/\sqrt2)&\operatorname{diag}(-2,24/7)&\text{saddle points}
\end{array}
$$
In particular the two <local minima> are also <global minima>: the squared term is nonnegative and $\ln(3+y^2)\ge\ln3$, with equality in both bounds exactly at $(\pm1,0)$. The <local maximum> at the origin is not a <global maximum>, since $f$ grows without bound as $x^2+y^2\to\infty$. The opposite diagonal signs at the other two points establish their <saddle point> character.
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