Finding critical points of f(x,y) Find the critical point of $$ f(x,y) = 3x^3 + 3y^3 + x^3y^3 $$
To do this, I know that I need to set $$f_y = 0, f_x = 0 $$
So $$f_x= 9x^2 + 3x^2y^3$$
$$f_y = 9y^2 + 3y^2x^3$$
Then you solve for x,  but substituting these two equations into each other.
But somehow I ended up with $$x = y$$ and thats not very helpful.
Is there something I did wrong or misunderstood?
 A: Well, $f_x=0$ is $3x^2(3-y^3)=0$, so either $x=0$ or $y^3=3$ (among reals it's $\sqrt[3]3$).
And, at the same time $f_y=0$ must also hold, that is ($y=0$ or $x=\sqrt[3]3$).
It gives you the $(0,0)$ and $(\sqrt[3]3,\sqrt[3]3)$ solutions.
A: If you solve the system $f_x=0=f_y$ , you will the following four points: 
         [0, -1.44, -9], [-1.44, 0, -9], [0, 0, 0], [-1.44, -1.44, -9] 

These are all critical points. Now, we should examine the signs of $f_{xx}$ and $$f_{xx}f_{yy}-f^2_{xy}$$ at these points. If the Hessian gets zero at any of them, then we don't know whether that point is max, min or a saddle point and so we need additional considerations.

A: Getting $x=y$ is very useful! This is because say you have two equations:
$$1. f_x=9x^2+3x^2y^3$$
$$2. f_y=9y^2+3y^2x^3$$
Substituting $x=y$, or $y=x$ into both equations and making the left side equal to zero will yield the same result:
$$0=9x^2+3x^2x^3$$
$$0=9x^2+3x^5$$
$$0=3x^2(3+x^3)$$
$$(0=3x^2) or (3+x^3=0)$$
$$(0=x^2) or (x^3=-3)$$
$$(x=0) or (x=-(3)^{1/3})$$
Again, since we know that $x=y$, we know that the critical points are $(0,0,0)$ and $(-3^{1/3},-3^{1/3},-9)$
A: The two equations
$$f_x=3x^2(3+y^3)=0,\qquad f_y=3y^2(3+x^3)=0$$
lead to the two critical points $(0,0)$, $\>(-\sqrt[3]3,-\sqrt[3]3)$.
In order to qualify the critical point $(0,0)$ we consider the function $\phi(x):=f(x,0)=3x^3$. Since $\phi$ assumes as well positive as negative values in the immediate neighborhood of $0$ we can conclude that $f$ does not assume a local extremum at $(0,0)$.
To analyze the critical point $(-\sqrt[3]3,-\sqrt[3]3)$ we compute the Hessian
$$\left[\matrix{18x+6xy^3 &9x^2y^2\cr 9x^2y^2 &18y+6yx^3\cr}\right]\ .$$
Its determinant is
$$9xy\bigl(36+12(x^3+y^3)-5x^3y^3\bigr)\ ,$$
which is negative at $(-\sqrt[3]3,-\sqrt[3]3)$. Therefore we don't have a local extremum at $(-\sqrt[3]3,-\sqrt[3]3)$ either.
