If you have a boundary which is given by the level curve of a function (here $x^{2} + y^{2}$), you can use the Lagrange multiplier theorem (see here). You have to check the points of the boundary such that the gradient of the function you want to study and the gradient of the function giving the domain are parallel.
In this case you have that the only critical point in $\mathbb{R}^{2}$ is $\left(-1 , -1/2 \right)$. $f$ is the sum of two quadratic functions in $x$ adn $y$ separately and with coefficients of $x^{2}$ and $y^{2}$ both positive. Hence $\mathrm{lim}_{||\left(x,y\right)|| \to +\infty}f\left(x,y\right)=+\infty$, so the point we have is the absolute minimum, and it lies in the circle. Now to find the maximum in the circle check the points of the circle of radius 2 such that $\left(2x+2,2y+1\right)$ is parallel to $\left(2x,2y\right)$. If you put them in a matrix and want the determinant to be $0$, it gives you the condition $2y=x$.
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
You can reduce $f_x, f_y, f_z$ by dividing out the $3$ to:
$$x^2-3y-3z+9 = 0 \\ y^2-3x = 0 \\z^2-3x =0$$
Update
From $z = \pm y$, we substitute in the first equation and have:
$$x^2 - 3 y+ 3 (\pm~ y) + 9 = 0$$
This gives two cases to check:
$$x^2 - 6y + 9 = 0, x^2 + 9 = 0$$
The second does not work, for the first, we know that $x = \dfrac{1}{3} y^2$. Substitute and you get $y = 3, 1.63107$ and two imaginary roots that we can immediately ignore (real numbers only). Now use those two $y$ values, substitute to find the corresponding $x$ and $z$ values.
One of those has a solution $x = 0.886793$ and another solution $x = 3$. Find the corresponding $z$ values. All other choices/approaches lead to these roots and other imaginary roots that we can ignore.
So, we have the following roots:
$$(x, y, z) = (3,3,3) ~~\mbox{or} ~~ (0.886793, 1.63107, 1.63107)$$