Let's see. First we need to compute the points where the partial derivatives of the function are zero. The partial derivatives are:
$$
d f(x,y) = (4x-4, 2y - 4)
$$
Which is the zero vector when $x=1$ and $y=2$.
To see whether this point is a maximum, a minimum or a saddle point we can compute the Hessian matrix of second derivatives.
$$
H_f =
\left( \begin{array}{ccc}
4 & 0 \\
0 & 2
\end{array} \right)
$$
As $H_f$ is definite positive, we conclude that there is a local minimum in $x=1$, $y=2$.
Now we have to study the boundary region for maximum and minimums.
Using the functions that you worked out to define the edges of the triangle, we can evaluate the function there.
$$
f_1(x)=f(x,2)=2x^2-4x-3\\
f_2(x)=f(0,y)=y^2-4y+1\\
f_3(x)=f(x,2x)=6x^2-12x+1
$$
Find the maximum and minimums of this functions in the area which falls within the region of interest and you will have found every local maximum and minimum.
From your calculation, you have shown that there is no critical point in the interior of the half-disk, but the left endpoint of the semicircular boundary is one. The index of that point is
$$ D \ \ = \ \ f_{xx}·f_{yy} \ - \ (f_{xy})^2 \ \ = \ \ 2·(-2) \ - \ 0^2 \ = \ -4 \ \ < \ \ 0 \ \ , $$
which identifies $ \ (-1 \ , \ 0 ) \ $ as a saddle-point. This makes sense since the function is "concave upward" in the $ \ x-$ direction about $ \ x \ = \ -1 \ $ , due to the $ \ x^2 \ + \ 2x \ = \ (x + 1)^2 \ - \ 1 \ $ terms, but is "concave downward" in the $ \ y-$ direction because of the $ \ -y^2 \ $ term. As we will need it for comparisons later, we compute that $ f(-1,0) \ = \ (-1)^2 \ + \ 2(-1) \ - \ 0^2 \ = \ -1 \ \ . $
The lower boundary of the region is the $ \ x-$ axis over the interval $ \ [ \ -1 \ , \ 1 \ ] \ $ . Inserting $ \ y \ = \ 0 \ $ into the function expression gives $ \ f(x,0) \ = \ (x + 1)^2 \ - \ 1 \ \ . $ We have already found $ \ f(-1,0) \ $ to be the minimum on this line segment; the right endpoint has the value $ f(1,0) \ = \ (1+1)^2 \ - \ 1 \ = \ 3 \ \ . $
It remains to examine the upper semicircular boundary. There are a number of ways we might set this up, but to avoid having to work with a square-root or trigonometric identities, we will use the circle equation to write $ \ y^2 \ = \ 1 \ - \ x^2 \ $ and express our function on this section of the boundary as
$$ \ \phi(x) \ = \ (x + 1)^2 \ - \ 1 \ - \ (1 - x^2) \ \ = \ \ (x + 1)^2 \ + \ x^2 \ - \ 2 \ \ . $$
[This gives us the correct values for the function at the endpoints of the region:
$$ \phi(-1) \ \ = \ \ ([-1] + 1)^2 \ + \ (-1)^2 \ - \ 2 \ \ = \ -1 \ \ \ , \ \ \ \phi(1) \ \ = \ \ (1 + 1)^2 \ + \ 1^2 \ - \ 2 \ \ = \ \ 3 \ \ . \ ] $$
We find $ \ \phi'(x) \ = \ 2·(x + 1) \ + \ 2x \ \ = \ \ 0 \ \ \ \Rightarrow \ \ \ 4x \ \ = \ -2 \ \ \ \Rightarrow \ \ \ x \ = \ -\frac{1}{2} \ \ , $ for which $ \phi \left( -\frac{1}{2} \right) \ \ = \ \ \left( \ \left[-\frac{1}{2} \right] + 1 \ \right)^2 \ + \ \left(-\frac{1}{2} \right)^2 \ - \ 2 \ \ = \ \ \frac{1}{4} \ + \ \frac{1}{4} \ - \ 2 \ \ = \ -\frac{3}{2} \ \ . $
Hence, we observe that the absolute maximum of the function on this closed half-disc is $ f(1,0) \ = \ 3 \ $ and the absolute minimum is $ f \left( -\frac{1}{2} \ , \ \frac{\sqrt{3}}{2} \right) \ = \ -\frac{3}{2} \ \ . $ The saddle point at $ \ (-1 \ , \ 0 ) \ $ has no special role in this region.
Best Answer
Considered the regularity of the objective function, any local or global extrema will occur either in stationary points in the interior of the region, or at points in the boundary.
Setting $y=-1$, you see that there is no hope of having a global minimum.
At the boundary, we have $f = -6y^2, y\ge -1$ or $f = -8 x^3 -12 x -6, x \ge 0.$
At the only local maximum in the interior, we have $f = \frac 12$.
Putting all this together, there is no global minimum and either the global maximum is $\frac 12$, or there is no global maximum. Can you proceed?