= Solution
The <gradient> equations factor as
$$
f_x=6(x^2+y^2-25)=0,\qquad f_y=3y(4x-3y)=0.
$$
For $y=0$ they give $x=\pm5$. Otherwise $y=4x/3$, which gives $x=\pm3$ and the matching $y=\pm4$. Thus there are exactly four <stationary points>.
The <Hessian matrix> is
$$
H_f=\begin{pmatrix}12x&12y\\12y&12x-18y\end{pmatrix}.
$$
At $(5,0)$ it is $60I$, and at $(-5,0)$ it is $-60I$. At $(3,4)$ and $(-3,-4)$ its <determinant> is $-3600<0$. The <second-derivative test> therefore yields
$$
\boxed{\begin{array}{c|c|c}
\text{point}&\text{type}&f\text{ value}\\\hline
(5,0)&\text{strict local minimum}&-500\\
(-5,0)&\text{strict local maximum}&500\\
(3,4)&\text{saddle point}&-300\\
(-3,-4)&\text{saddle point}&300.
\end{array}}
$$
The extrema are local: along $y=0$ the cubic is unbounded both above and below.
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