I am looking for the extrema of a function $G(y_1,y_2,y_3,y_4)$ subject to the constraint $y_1 = y_4 + y_2y_3.$ We know that $G$ is defined if $(y_2,y_3,y_4)$ is in the cylinder $\mathbb{D} \times \mathbb{R},$ where $\mathbb{D}$ is the open unit disk of the plane. At a stationary point $(y^*,\lambda^*)$ of the usual Lagrangian $\mathcal{L}(y, \lambda),$ the Hessian is $$ \nabla_{yy} \mathcal{L}(y^*,\lambda^*) = \begin{bmatrix} 4 & 0 & 0 & -2 \\ 0 &-2({y_4^*}^2+1) & 2y_4^* & 0 \\ 0& 2y_4^* & -2{y_4^*}^2 &0 \\ -2&0&0&0 \end{bmatrix}.$$
Now, set $w^* = [w_1,w_2,w_3,w_1] \in \mathbb{R}^4,$ which is in the tangent plane of the constraint. Then
$$ w^* \nabla_{yy} \mathcal{L}(y^*,\lambda^*) w < 0$$
if $y_4^* \neq 0 $ and $ w_2,w_3 \neq 0.$
Here is the messy thing for me: Now consider the $y_4^*$-section of the cylinder $\mathbb{D} \times \mathbb{R};$ that is, the set of $(y_2,y_3,y_4^*)$-s. The previous inequality of the Hessian means that the restriction of $G$ onto the set $(y_1,y_2,y_3,y_4^*)$ (subject to the constraint above) has strict local maximum at $y^*$? Any hint is welcomed.
Even restricted, $y^*$ need not be a local maximum. Necessary would be $$ w^* \nabla_{yy} \mathcal L(y^*, \lambda^*) w \le 0 $$ for all $w^* = (w_1, w_2, w_3, 0)$ in the tangent space, that is $$ w_1 - y_3^*w_2 - y_2^* w_3 = 0. $$ Strictly less for sufficient condition. Since the (1,1) element of the hessian is $1$, it depends on $y_2^*$ and $y_3^*$.