Calling
$$
f(x,y,z) = z^2 - 2 x^2 - y^2 - 4 x y - 2 x z - z + x\\
g_1(x,y,z) = 6 x^2 + y^2 + z^2 + 4 x y - 2 z x - 1\\
g_2(x,y,z) = 4 x^2 + y^2 + 4 x y + 2 x z - z^2
$$
and introducing the auxiliary slack variables $\epsilon_i$ to transform the inequalities into equalities, we have the lagrangian
$$
L(X,\lambda,\epsilon) = f(x,y,z)+\lambda_1(g_1(x,y,z)+\epsilon_1^2)+\lambda_2(g_2(x,y,z)+\epsilon_2^2)
$$
The stationary points are obtained by solving
$$
\left\{
\begin{array}{rcl}
-4 x-4 y+\lambda_1 (12 x+4 y-2 z)-2 z+\lambda_2(8 x+4 y+2 z)+1&=&0 \\
-4 x-2 y+\lambda_1(4 x+2 y)+\lambda_2(4 x+2 y)&=&0 \\
-2 x+\lambda_2 (2 x-2 z)+2 z+\lambda_1(2 z-2 x)-1&=&0 \\
\epsilon_1^2+6 x^2+y^2+z^2+4 x y-2 x z-1&=&0 \\
\epsilon_2^2+4 x^2+y^2-z^2+4 x y+2 x z&=&0 \\
2 \epsilon_1\lambda_1&=&0 \\
2 \epsilon_2 \lambda_2&=&0 \\
\end{array}
\right.
$$
with the solutions
$$
\begin{array}{cccccccc}
x & y & z & \lambda_1 & \lambda_2 & \epsilon_1 & \epsilon_2 & f(x,y, z) \\
-\frac{1}{4} & \frac{1}{2} & 0 & 0 & -1 & \frac{\sqrt{\frac{7}{2}}}{2} & 0 & -\frac{1}{8} \\
0 & 0 & -1 & -\frac{3}{2} & 0 & 0 & 1 & 2 \\
0 & 0 & \frac{1}{2} & 0 & 0 & \frac{\sqrt{3}}{2} & \frac{1}{2} & -\frac{1}{4} \\
0 & 0 & 1 & -\frac{1}{2} & 0 & 0 & 1 & 0 \\
0 & -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2} \left(2+\frac{1}{\sqrt{2}}\right) & 0 & 0 & \frac{1}{\sqrt{2}}
\\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & -\frac{1}{2 \sqrt{2}} & \frac{1}{2} \left(2+\frac{1}{\sqrt{2}}\right) & 0 & 0 & \frac{1}{\sqrt{2}} \\
0 & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2} \left(2-\frac{1}{\sqrt{2}}\right) & 0 & 0 & -\frac{1}{\sqrt{2}} \\
0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & \frac{1}{2 \sqrt{2}} & \frac{1}{2} \left(2-\frac{1}{\sqrt{2}}\right) & 0 & 0 & -\frac{1}{\sqrt{2}} \\
\frac{1}{4} & -\frac{1}{2} & \frac{1}{2} & 0 & -1 & \frac{\sqrt{\frac{7}{2}}}{2} & 0 & -\frac{1}{8} \\
-\frac{1}{\sqrt{2}} & \sqrt{2} & 0 & \frac{1}{2} \left(-2+\frac{1}{\sqrt{2}}\right) & -\frac{1}{2 \sqrt{2}} & 0 & 0 & 1-\frac{1}{\sqrt{2}} \\
-\frac{1}{\sqrt{2}} & \sqrt{2} & -\sqrt{2} & \frac{1}{2} \left(-2-\frac{1}{\sqrt{2}}\right) & \frac{1}{2 \sqrt{2}} & 0 & 0 &
1+\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\sqrt{2} & 0 & \frac{1}{2} \left(-2-\frac{1}{\sqrt{2}}\right) & \frac{1}{2 \sqrt{2}} & 0 & 0 & 1+\frac{1}{\sqrt{2}} \\
\frac{1}{\sqrt{2}} & -\sqrt{2} & \sqrt{2} & \frac{1}{2} \left(-2+\frac{1}{\sqrt{2}}\right) & -\frac{1}{2 \sqrt{2}} & 0 & 0 &
1-\frac{1}{\sqrt{2}}\\
\end{array}
$$
as we can observe, there are a lot of stationary points. Stationary points with at least one $\epsilon_i = 0$ are located at the feasible region boundary. Solutions with all $\epsilon_i \ne 0$ are internal stationary points.
Best Answer
$f(x,y)=e^1e^{-2y^2}$, $0\le y^2\le1;$
$\max f(x,y) =e^1$, for $y=0,$ $(x=?);$
$\min f(x,y) =e^{-1}$, for $y=\pm 1,$ $(x=?).$