= Solution
For a differentiable equality constraint, the <Lagrange multiplier> method finds candidate <extrema> by solving
$$
\nabla f=\lambda\nabla g,\qquad g=c.
$$
The condition requires $\nabla g\ne0$ at the point: every tangent direction to the constraint is perpendicular to $\nabla g$, and at a <extremum> the directional <derivative> of $f$ vanishes in every such direction. Thus $\nabla f$ is parallel to $\nabla g$. One must then classify candidates and check any singular or boundary cases.
For the <ellipsoid>, its constraint <gradient> never vanishes on the surface, and its <compactness> guarantees a maximum and a minimum. The <Lagrange multiplier> equations become
$$
y=\frac{2\lambda x}{a^2},\qquad x=\frac{2\lambda y}{b^2},\qquad0=\frac{2\lambda z}{c^2}.
$$
If $\lambda=0$, then $x=y=0$ and $z=\pm c$, giving value $0$. If $\lambda\ne0$, then $z=0$; neither $x$ nor $y$ can vanish on this part of the surface. Combining the first two equations gives $4\lambda^2=a^2b^2$, and then $y=\pm(b/a)x$. The constraint yields $x^2=a^2/2$ and $y^2=b^2/2$. Therefore
$$
\boxed{\max xy=\frac{ab}{2},\qquad\min xy=-\frac{ab}{2}.}
$$
The maximum occurs at $(a/\sqrt2,b/\sqrt2,0)$ and its negative; the minimum at $(a/\sqrt2,-b/\sqrt2,0)$ and its negative. To verify these are the global <extrema>, use
$$
\frac{2|xy|}{ab}\leq\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq1.
$$
Equality is attained at exactly the points just found. The remaining two candidates have value $0$ between the extremal values.
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