$f(x,y,z)=x^3+y^3+z^3-3xy-3yz$
According to wolfram alpha, $f(0,0,0)$ is not a extreme value.
I considered why $f(0,0,0)$ is not a extreme value.
Is this reason correct?
When $x$ is very close to $0$ with $y=x, z=x$, $f(x,y,z)=f(x,x,x)=3x^2(x-2)<0=f(0,0,0)$
When $x$ is very close to $0$ with $y=-x, z=x$, $f(x,y,z)=f(x,-x,x)=x^2(x+6)>0=f(0,0,0)$
Therefore, on any point which are close to $(0,0,0)$, $(x_0,y_0,z_0)$ such that $f(x_0,y_0,z_0)<f(0,0,0)$ exists and $(x_0 ',y_0 ',z_0 ')$ such that $f(x_0 ',y_0 ',z_0 ')>f(0,0,0)$ exists.
Thus, $f(0,0,0)$ is not a extreme value.