This was billed as a "multivariable calculus" exercise, but it is unlike any calc exercise I have ever seen:
Exercise 26.3. The function $f(x,y)=8x^2+6xy+4x^3+3xy^2$ appears in Example 1.2.3, where its critical points are found: $(0,0)$, $(0,-2)$, $(-3/2,-1)$, $(1/6,-1)$.
(a) Compute the Hessian matrix at each of these points and for each determine if it is a local maximum, local minimum, or a saddle point.
(b) Sketch a contour plot of $f$ near each of the points $(-3/2,-1)$ and $(1/6,-1)$ by computing eigenvalues for the Hessian there. (It only matters to sketch approximate ellipses or hyperbolas aligned with the appropriate perpendicular lines through the critical point, indicating the longer axis direction in the ellipse case and the eigenline to which the asymptotes are "closer" in the hyperbola case.)
Part (a) is easy enough: the Hessian matrix is comes out to be $$ H_f=\begin{bmatrix}16+24x&6+6y \\ 6+6y&6y \end{bmatrix} $$ whence we have $$ H_f(0,0) =\begin{bmatrix}16&6 \\ 6&0 \end{bmatrix}, \;\;\;\; H_f(0,-2) =\begin{bmatrix}16&-6 \\ -6&-12 \end{bmatrix}, $$ $$ H_f(-3/2,-1) =\begin{bmatrix}-20&0 \\ 0&-6 \end{bmatrix},\;\;\;\;\; H_f(1/6,-1) =\begin{bmatrix}20&0 \\ 0&-6 \end{bmatrix}. $$ The second derivative test can then be applied to these points by taking the determinant of the Hessian: $$ |H_f(0,0)|=-36, \;\;\;\;|H_f(0,-2)|=-228, $$ $$ |H_f(-3/2,-1)|=120,\;\;\;\;|H_f(1/6,-1)|=-120 $$ so that $(0,0)$, $(0,-2)$, and $(1/6,-1)$ are saddle points, and $(-3/2,-1)$ is a local maximum.
For part (b), the eigenvalues of $H_f(-3/2,-1)$ are just the diagonal elements $-20$ and $-6$, whereas the eigenvalues of $H_f(1/6,-1)$ are $20$ and $-6$.
But what on earth does it mean to "sketch a contour plot" near those points, and why do we care about the eigenvalues of the Hessian?
Thanks!