Critical points of $f\left(x,y,z\right)=x^3+y^3+z^2+2xyz$ Find the critical points of the function and specify the nature of these points.
$$f:\mathbb{R}^3\rightarrow \mathbb{R}$$
$$f\left(x,y,z\right)=x^3+y^3+z^2+2xyz$$
So I solved the system of partial derivatives and I get $\left(0,0,0\right);\:\left(\frac{3}{2},\frac{3}{2},-\frac{9}{4}\right)$ as critical points.
But when I try to compute the Hessian Matrix I get the determinant equal to 0 for the point $(0,0,0)$. For my other point I get a negative determinant, which would mean a saddle point. How do I determine the nature of my first point though, since the second derivative test is inconclusive?
 A: You found correctly the critical points:
$$
\begin{cases}
3x^2+2yz=0 \\[4px]
3y^2+2xz=0 \\[4px]
2z+2xy=0
\end{cases}
$$
gives $z=-xy$ and so the first and second equations become
$$
\begin{cases}
3x^2-2xy^2=0 \\[4px]
3y^2-2x^2y=0
\end{cases}
$$
If $x=0$, also $y=0$ and $z=0$. If $y=0$, then also $x=0$ and $z=0$. Assuming $x\ne0$ and $y\ne0$, we get $x=3/2$, $y=3/2$ and $z=-9/4$.
Indeed the Hessian matrix at $(0,0,0)$ has zero determinant, so it cannot be used for determining the character of this stationary point. However, you can look at what happens on particular restrictions: consider the line
$$
\begin{cases}
x=at\\[4px]
y=bt\\[4px]
z=ct
\end{cases}
$$
that leads us to study
$$
g(t)=f(at,bt,ct)=(a^3+b^3+2abc)t^3+c^2t^2
$$
If $a=1$, $b=0$, $c=0$, we have $g(t)=t^3$.
Thus the origin is neither a point of maximum nor of minimum.
A: EDIT: Re-wrote my answer for the more general case. For a single-variable function, according to the following Theorem:
Suppose that $$f'(X)=f''(X)=...=f^{(n-1)}(X)=0,$$ but $f^{(n)}(X)\neq0$, and that $f^{(n)}(X)$ is continuous in some neighborhood of X, where $n\geq2$. If n is even, then f has a local maximum at X if $f^{(n)}(X)<0$ and a local minimum if $f^{(n)}(X)>0$. If n is odd, then f has a horizontal inflection point at X.
We can test this on the function $f(x)=x^3$, which has a critical point at $x=0$. The derivatives at this point are $f'(0)=0$, $f''(0)=0$, $f'''(0)=6$. In this case, $n=3$, which is odd, therefore according to the theorem we have a horizontal inflection point.
A: "A critical point with a second derivative equal to zero is neither a minimum or a maximum. It is called an "inflection point.""
No.  Consider the function $y= x^4$.  First, second, and third derivatives are all 0 at x= 0 but the function clearly has a minimum at x= 0.
An "inflection point" is a point where the second derivative changes sign.  The second derivative being 0 there is obviously a necessary condition but not a necessary condition.
