Let $h: \{(v,w,x,y) \in \mathbb{R}^4 : w <0 \} \to \mathbb{R}$ with $h(v,w,x,y) := 6v^2-12v+arctan(w)- \frac{1}{2}w+\exp(x^2)+x^2+y^2+\frac{1}{4}xy$
How can one find the criticial points, i.e. the local/global maxima and minima and saddle points of this function?
I know that a local maxima/minima $x_E$ of a function is its Zero of its derivative $f'(x_E) = 0$.
If $f''(x_E) > 0$ then there's a local minimum.
If $f''(x_E) < 0$ then there's a local maximum.
And if $f''(x_E) = 0$ we can't tell anything.
I don't know how to derivate the function twice, because of the condition that $w <0$ and how I should proceed afterwards.
The equilibrium points will be where the gradient $\nabla h = (h_v,h_w,h_x,h_y)$ is the $0$ vector. To tell whether or not it is a maximum/minimum/saddle, you have to look at the eigenvalues of the Hessian matrix, $$ H = \begin{pmatrix} h_{vv} & h_{vw} & h_{vx} & h_{vy} \\ h_{wv} & h_{ww} & h_{wx} & h_{wy} \\ h_{xv} & h_{xw} & h_{xv} & h_{xy} \\ h_{yv} & h_{yw} & h_{yx} & h_{yy} \end{pmatrix}. $$ By the equality of mixed partials the eigenvalues are real and the equilibrium point will be a minimum if each eigenvalue is positive, a maximum if they are all negative, and a saddle if they have different signs. $0$ eigenvalue is a still an indeterminate case.
You will also have to be careful with your domain. Since it does not contain the hyperplane $w=0$, it is possible that the extrema will lie on this hyperplane but your function cannot reach it in this domain.