Local Extrema of a 2D Surface in R^3 (Math GRE Question)

Question

The surface in question is $f(x,y) = x^3 + y^3 + 3xy$

I'm told that there are 2 distinct points $P$ and $Q$ of $f$ at which $f$ has a local maximum and a saddle point, respectively. How can I show the existence of these two points?

Here is what I've tried:
I try to find local extrema by finding the points at which the derivative of $f$ is $0$.

$f'(x,y) = (3x^2 + 3y, 3y^2 + 3x)$ and after some work it's apparent that the only local extrema lie on the line $y = (1-x)$.
I continue by calculating the Hessian matrix:
| 6x, 3|
| 3, 6y| =

|6x, 3|
|3, 6(1-x)| (if the derivative is to be 0)

By examining the trace, I know that the sum of the eigenvalues is 6. By examining the determinant, I know that the product of the eigenvalues is $36x(1-x) - 9$, which has a root at $1/2$ but is otherwise negative. So at $(1/2, 1/2)$ I have the eigenvalues $6, 0$ which means that I have a local minimum! And at all other points I have saddle points! So I must have done something wrong.

Aside: By symmetry of $f$ around the line $x=y$, I deduce that at least one of the points $P$ or $Q$ has a counterpart $P'$ or $Q'$ by reflection across $x=y$. Therefore the question must mean that there are at least 2 distinct points $P$ and $Q$.

This question comes from http://www.wmich.edu/mathclub/files/GR8767.pdf #55

Answer 1

Find the extrema by solving the partials simultaneously being zero. In this case, $3x^2+3y=0$ and $3y^2+3x=0$. Subtracting, $$3(x^2-y^2)-3(x-y)=0\quad\longrightarrow\quad (x-y)(x+y-1)=0.$$ If $x=y$ then $x^2=-x$ implies $x=y=0$ or $x=y=-1$. If $y=1-x$ then $x^2-x+1=0$ which has no real solutions.

Therefore the two extrema are $(0,0)$ and $(-1,-1)$. Evaluate the Hessian at both points to determine the nature of the extrema.

Answer 2

I doubt if your computation is correct. Can you double check? Even the first step $3x^{2}+3y=0,3y^{2}+3x=0$ give rise to $y=(1-x)$ seem to be wrong. You can only show $x=-y^{2},y=-x^{2}$. The fact $x=-x^{4}$ showed $1+x^{3}=0$ or $x=0$.

But we do not have complex numbers available. So both $x$ and $y$ must be $-1$ or $0$ for the local extrema to hold. The Hessian matrix has the form $(6x,3),(3,6y)$ and the determinant is $36xy-9$. Use the second derivative test we conclude at $(-1,-1)$ we had a local maximum and at $0$ we had a saddle point.