The function is $f(x_1,x_2,x_3) = e^{x_1^2}+x_2^4+x_2^2x_3^2+x_3^4+6x_2+6x_3$
The gradient is $∇f(x)=[2x_1e^{x_1^2} ,4x_2^3+2x_2x_3^2+6, 4x_3^3+2x_3x_2^2+6]$
Now to find the extremum points the gradient must be zero and the only point I could find was $x=(0,-1,-1),$ however I'm not sure it is the only extremum point.
Furthermore, the $∇^2f(x)= \begin{pmatrix} 2e^{x_1^2}+4x_1^2e^{x_1^2} & 0 & 0 \\ 0 & 12x_2^2+2x_3^2 & 4x_2x_3 \\ 0 & 4x_3x_2 & 12x_3^2+2x_2^2 \end{pmatrix}$
There's a theory that states if $∇^2f(x)≥0$ then the function is convex, therefore any local extremum points are global.
At the point $x=(0,-1,-1)$ we do have that $∇^2f(x)≥0$, since all the leading principle minors are positive.
Does this mean that the function has one extremum point and that it's a global one?
It is correct, there is only one extremum point, the local minimum is a global minimum and is the only local extremum of the function.