We compute the partial derivatives:
$\frac{\partial f}{\partial x} = 6x^2$
$\frac{\partial f}{\partial y} = 4y^3$
Setting these to zero, we get $x = 0$ and $y = 0$. Thus, the only critical point is $(0, 0)$, which is inside the region.
At $(0, 0)$, $f(0, 0) = 0$.
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