You should think geometrically. The function $$f(x,y)=-x^{2012}-y^{2012}$$ can be written $$f(x,y)=-(x^{2012}+y^{2012})$$ But because 2012 is even, $x^{2012}\geq0$ and $y^{2012}\geq 0$. Therefore $x^{2012}+y^{2012}\geq0$ and so $f(x,y)\leq 0$ for all points $(x,y)$. Therefore your critical point $(0,0)$ is a global maximum.
Whoever assigned this problem probably wanted you to notice that your function $f$ is similar to
$$g(x,y)=x^2+y^2$$which is an upward-opening elliptic paraboloid that has a global minimum at the origin. Similarly, $$h(x,y)=-x^2-y^2$$ is a downward-opening elliptic paraboloid that has a global maximum at the origin.
The only difference between these functions and your function $f$ is that $f$ rises much, much faster when $|x|>1$ and $|y|>1$ and is much, much flatter when $|x|<1$ and $|y|<1$, because the exponent is so much higher. In fact, the graph would look more like a box than a paraboloid because the walls of the surface are so steep.
In general, there's no surefire method for analyzing the local behavior of functions where the second derivative test comes back inconclusive. In practice, you should think geometrically or look at higher order derivatives to get a sense of what's going on.
To use the latter approach, consider taking the 2012th partial derivatives of your function. Both the $x$ and $y$ partial at the origin will be $-2012!$, and combining this with the multivariate Taylor expansion (and reflecting on the mixed partials) gives another way of seeing that your function is negative definite near the origin. (The reason the second derivative test fails for this function is that it is too flat near its critical point. This extreme flatness is what makes so many of the higher-order derivatives zero.)
But your function is so simple to understand that its global properties are obvious if you think geometrically. Other functions might require you to think harder about what's happening graphically, in which case the analytic approach with higher derivatives could be helpful.
After setting all three equations to $0$, multiply the first equation by $z$, the second by $x$, and the third by $xy$ to get:
\begin{align}
2xz + 2xyz &= 0 \quad (1) \\
x^3 + 2xyz &= 0 \quad (2) \\
xy^3 + 2xyz - 4xy &= 0 \quad (3) \\
\end{align}
\begin{align}
(1) - (2) &\Rightarrow 2xz - x^3 = 0 &(4) \\
(2) - (3) &\Rightarrow x^3 - xy^3 -4xy = 0 &(5) \\
(1) - (3) &\Rightarrow 2xz - xy^3 -4xy = 0 &(6) \\
\end{align}
Subtracting $(6)$ from $(5)$ gives $x^3 - 2xz = x(x^2-2z) = 0$ so that $x = 0$ or $z = x^2/2$.
In the first case, we get $2yz=0$ from the second equation, so either $y=0$ or $z=0$.
If $y=0$, then $2z-4=0$ so that $z=2$, which gives the solution $(0,0,2)$.
If $z=0$ then $y^2 = 4$ so that $y=\pm2$, which gives the solutions $(0,2,0)$, and $(0,-2,0)$.
In the second case, we get $x\neq0$ so that $z=x^2$.
Then, first original equation gives $2x(1+y) = 0$. But $x \neq 0$, so we must have $y=-1$.
Plug this into the third original equation to get $(-1)^2 + 2z - 4 = 0$, i.e. $z = \frac{3}{2}$.
Now plug $y=-1$ and $z=3/2$ into the second original equation to get $x^2 +2(-1)(\frac{3}{2}) = 0$ which gives $x^2 = 3$ or $x = \pm\sqrt 3$.
This gives the final solutions, $(-\sqrt 3, -1, \frac{3}{2})$, and $(\sqrt 3, -1, \frac{3}{2})$.
So there are three solutions: $(0, 0, 2)$, $(0,-2,0)$, $(0, 2, 0)$, $(-\sqrt 3, -1, \frac{3}{2})$, and $(\sqrt 3, -1, \frac{3}{2})$. You can verify that these work by plugging them in to the original equations.
Best Answer
You don't "need" bordered Hessians. You can do as follows (which I suppose is equivalent to bordered Hessian, but also works with inequality constraints). But as discussed below, you can't just use the definiteness of the unmodified (full) Hessian of the objective function to ascertain the nature of a stationary point if any constraints are active (satisfied with equality) at that point.
Presuming the objective function and all constraint functions are twice continuously differentiable, then at any local minimum, the Hessian of the Lagrangian projected into the null space of the Jacobian of active constraints must be positive semidefinite. I.e., if $Z$ is a basis for the null space of the Jacobian of active constraints, then $Z^T*(\text{Hessian of Lagrangian})*Z$ must be positive semidefinite. This must be positive definite for a strict local minimum.
So the Hessian of the objective function in a constrained problem having active constraint(s) need not be positive semidefinite if there are active constraints.
All the preceding applies to local maximum (strict local maximum), if positive semidefinite is changed to negative semidefinite (negative definite).
Notes:
1) Active constraints consist of all equality constraints, plus inequality constraints which are satisfied with equality.
2) See the definition of the Lagrangian at https://www.encyclopediaofmath.org/index.php/Karush-Kuhn-Tucker_conditions .
3) If all constraints are linear, then the Hessian of the Lagrangian = Hessian of the objective function because the 2nd derivatives of linear functions are zero. But you still need to do the projection jazz if any of these constraints are active. Note that lower or upper bound constraints are particular cases of linear inequality constraints. If the only constraints which are active are bound constraints, the projection of the Hessian into the null space of the Jacobian of active constraints amounts to eliminating the rows and columns of the Hessian corresponding to those components on their bounds.
4) Because Lagrange multipliers of inactive constraints are zero, if there are no active constraints, the Hessian of the Lagrangian = the Hessian of the objective function, and the Identity matrix is a basis for the null space of the Jacobian of active constraints, which results in the simplification of the criterion being the familiar condition that the Hessian of the objective function be positive semidefinite at a local minimum (positive definite if a strict local minimum).