Hint: Look at the corresponding closed region $R$, so that the extrema will either be critical points in int$R$ or they will lie on the boundary of $R$. For the saddle points, use the second derivative test.
Second hint:
Step 1:
Solve the following equations and use the points in the second derivative test to decide if they are local extrema or saddle points. Suggestion: let $z=y/x$
$f_{x}=4x^{3}-4y=0\\f_{y}=4y^{3}-4x=0$
Step 2:
Now consider each piece of the boundary, making the appropriate substitution and extremize the function (of a single variable) you get in each case. For example, if we look at the top side of the square, where $y=2$ and $-2<x<2$, we get
$f(x,2)=x^{4}-8x+16$, from which $f'(x,2)=4x^{3}-8=0$ gives $x=2^{1/3}$ as a critical point. Now determine what type of critical point it is, substitute into $f(x,y)$ and compare it with what you got in step 1. Then check $x=2, x=-2$ in $f(x,2)$ and write down the result.
repeat for the other sides of the square.
All told, you get a table of values from which you can read off the maxima and minima.
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 also solve this by using Lagrangian multipliers. You have the problem $$ \underset{x,y}{\text{minimize}}\ \quad \quad x^2+3y^2-y\\ \text{subject to}\ \ x^2+2y^2-2\leq 0 $$ To put it in Lagrangian form, we simply restate the problem as $$ L(x,y,\lambda) = x^2+3y^2-y + \lambda(x^2+2y^2-2) $$ What we need to do now is to find all the KKT points. To do so, we solve the following problem $$ \frac{\partial L}{\partial x}=0 \iff 2x+2\lambda x=0\quad (1)\\ \frac{\partial L}{\partial y}=0 \iff 6y-1+4\lambda y=0\quad (2)\\ \lambda(x^2+2y^2-2)=0 \quad (3) $$ First, assume that $\lambda=0$, then we must have (from $(1)$ and $(2)$) $x=0,y=1/6$. This is our first KKT-point. To find the others, assume that $\lambda\neq 0$. Then from $(1)$ we get either $x=0$ or $\lambda=-1$ from $(2)$. This yields the points $(x,y)=(0,1)$, $(x,y)=(0,-1)$, and $(x,y) = (\frac{\pm\sqrt(3)}{2},\frac{1}{2})$. Now lets check which point is minimium/maximum: $$ f(0,1/6) = -\frac{1}{12}\\ f(0,1) = 2\\ f(0,-1) = 4\\ f(\frac{\sqrt(3)}{2},\frac{1}{2})=1\\ f(\frac{-\sqrt(3)}{2},\frac{1}{2})=1 $$ Thus the minimum is $-\frac{1}{12}$ attained at $(x,y) = (0,\frac{1}{6})$ and the maximum is $4$ attained at $(x,y) = (0,-1)$.
Note: If you had more constraints (equality or inequality) you could use the same approach, however then you would have more Lagrange multipliers. (3) is called complementary slackness.