Since the function was apparently altered after at least some of the answers were posted, I'll address a peculiarity of this problem that may have raised Ilda's original concern.
It should be remarked upon that while the "level curves" of the function $ \ f(x, \ y) \ = \ x^2 \ + \ 4xy \ + \ 4y^2 \ $ may give the impression of being (rotated) ellipses, this is in fact the equation of a degenerate conic: since $ \ x^2 \ + \ 4xy \ + \ 4y^2 \ = \ (x \ + \ 2y)^2 \ $ , a level curve is actually a pair of parallel lines. Moreover, since we are dealing only with real-valued coordinates, the minimum value of the function is zero, which "collapses" that level curve to a single line through the origin. This is the case quite independently of the constraint.
So a geometric interpretation of the Lagrange-multiplier problem is to find the value(s) of the function for which the degenerate conic is tangent to the "constraint ellipse" $ \ x^2 \ + \ 2y^2 \ = \ 4 \ $ . For simple convenience, and as a reminder of the possible values of the function, we will express the level curves of $ \ f(x, \ y) \ $ as $ \ (x \ + \ 2y)^2 \ = \ c^2 \ $ . Here is a graph of the situation:

Now, we can solve some of this problem without calculus, just for the purpose of investigating the nature of the extremization. If we describe the parallel lines by $ \ x \ + \ 2y \ = \ \pm c \ $ , we can locate any possible intersection points with the constraint ellipse from
$ \ ( \ \pm c \ - \ 2y \ )^2 \ + \ 2y^2 \ = \ 4 \ \ \Rightarrow \ \ 6y^2 \ \mp \ 4cy \ + \ (c^2 \ - \ 4) \ = \ 0 \ \ . $
The discriminant of this quadratic equation is $ \ ( \ \mp 4c \ )^2 \ - \ 4 \cdot 6 \ (c^2 \ - \ 4) \ = \ 96 \ - \ 8c^2 \ $ . Each line has a single intersection with the ellipse when this discriminant is zero, which occurs for $ \ c^2 \ = \ 12 \ $ . The discriminant is negative for larger values of $ \ c^2 \ $ , meaning that there would be no intersection points, so the maximum value of $ \ f(x, \ y) \ $ under the constraint is $ \ 12 \ $ . As we said above, the minimum value is necessarily zero, which does produce intersection points with the ellipse; these, however, are not tangent points.
[Side note -- with a little further work, we can determine the two tangent points for $ \ f(x, \ y) \ = \ 12 \ $ to be $ \ \left( \pm \frac{2}{3} \sqrt{3} , \ \pm \frac{2}{3} \sqrt{3} \right) \ $ , which are placed symmetrically about the origin, as we expect when both the function to be extremized and the constraint function are symmetrical under inversion (what is also called "symmetry about the origin"). Ordinarily in an optimization of this sort, as joriki observes, each of these points would be associated with the maximum or minimum of the function. We see something of that here if we write the equations of the parallel lines individually, but in our situation, those lines represent only one value of the function.]
We can now look at what the Lagrange-multiplier method leads us to. The Lagrange equations for (the final version of) the function under the constraint are
$ 2x \ + \ 4y \ = \ \lambda \cdot 2x \ \ , \ \ 4x \ + \ 8y \ = \ \lambda \cdot 4y \quad \mathbf{[1]} $
$ \Rightarrow \ \ ( \ 1 \ - \ \lambda \ ) \ x \ + \ 2y \ = \ 0 \ \ , \ \ x \ + \ ( \ 2 \ - \ \lambda \ ) \ y \ = \ 0 \ \ . \quad \mathbf{[2]} $
[this differs from Christian Blatter's answer, owing to the subsequent revision of the posted function]
We could also solve Equations [ 1 ] for
$ \lambda \ = \ \frac{x \ + \ 2y}{x} \ = \ \frac{x \ + \ 2y}{y} \ \ . $
For $ \ x \ \ne \ 0 \ , \ y \ \ne \ 0 \ $ , we may write
$ x^2 \ + \ 2xy \ = \ xy \ + \ 2y^2 \ \ \Rightarrow \ \ x^2 \ + \ xy \ - \ 2y^2 \ = \ 0 \ \ ; $
we can put this together with the fact that (since there is symmetry about the origin for $ \ f(x, \ y) \ $ and the constraint function) the extrema lie on lines through the origin $ \ y \ = \ mx \ $ to obtain
$ x^2 \ + \ mx^2 \ - \ 2 \ m^2 x^2 \ = \ 0 \ \ \Rightarrow \ \ x^2 \ ( 2m^2 \ - \ m \ - \ 1 ) \ = \ 0 $
$ \Rightarrow \ \ m \ = \ 1 \ \ , \ \ -\frac{1}{2} \ \ \text{for} \ \ x \ \ne \ 0 \ \ . $
From these slopes, we can determine
$ \mathbf{m = 1 :} \quad y \ = \ x \ \ \Rightarrow \ \ x^2 \ + \ 2x^2 \ = \ 4 \ \ \Rightarrow \ \ x^2 \ = \ \frac{4}{3} \ \ , $ $ f \left(\frac{4}{3} , \ \frac{4}{3} \right) \ = \ x^2 \ + \ 4x^2 \ + \ 4x^2 \ = \ 9 \ \cdot \frac{4}{3} \ = \ 12 \ ; \ \text{also} \ \ \lambda \ = \ \frac{x \ + \ 2x}{x} \ = \ 3 \ \ ; $
$ \mathbf{m = -\frac{1}{2} :} \quad x \ + \ 2y \ = \ 0 \ \ , \ \ f \left(x , \ -\frac{1}{2}x \right) \ = \ x^2 \ - \ 2x^2 \ + \ x^2 \ = \ 0 \ \ ; $ $ \lambda \ = \ \frac{x \ + \ 2y}{x} \ = \ \frac{x \ + \ 2y}{y} \ = \ 0 \ \ , \ \ \text{for} \ \ x, \ y \ \ne \ 0 \ \ . $
We obtain from this the extremal values of our function, and also show that the minimum value corresponds to $ \ \lambda \ = \ 0 \ $ , which does not have tangent points on the constraint ellipse.
We can arrive at this latter result in another way: Equations [ 2 ] become a linearly dependent system for $ \ \lambda \ = \ 0 \ $ , producing the single equation $ \ x \ + \ 2y \ = \ 0 \ $ with the same conclusions. The features we have seen here are not typical of an optimization problem involving functions related to non-degenerate conic sections.