Let, \begin{align} f(x,y)=\frac{1}{y^2}+\left(\frac{x}{y}+\sqrt{2-\frac{1-x^2}{y^2}}\right)^2, \end{align}
where $0 \le x\le 1$, $-1 \le y \le0$, $x^2 + y^2 \le 1$ and $x^2 + 2y^2 \ge 1$. What is the $\min\limits_{x,y}f(x,y)$ subject to these constraints?
$f(x,y)$ is not convex under these constraints because $0 \le x\le 1$, $-1 \le y \le0$, $x^2 + y^2 \le 1$ and $x^2 + 2y^2 \ge 1$ is not a convex set.
Given, \begin{align} f(x,y)=\frac{1}{y^2}+\left(\frac{x}{y}+\sqrt{2-\frac{1-x^2}{y^2}}\right)^2, \end{align}
where $0 \le x\le 1$, $-1 \le y \le0$, $x^2 + y^2 \le 1$ and $x^2 + 2y^2 \ge 1$.
We can find the $\min\limits_{x,y}f(x,y)$ when $\nabla f(x,y)=0$
Thus, we can expand $f(x,y)$ to become:
$$f(x,y) = 2 + \frac{2 x^2}{y^2} + \frac{2 x}{y}\sqrt{2 - \frac{1}{y^2} + \frac{x^2}{y^2}}$$
Then, the gradient becomes:
$$\nabla f(x,y) = \left( \frac{4xy\sqrt{\frac{x^2+2y^2-1}{y^2}}+4x^2+4y^2-2}{y^3\sqrt{\frac{x^2+2y^2-1}{y^2}}}, \frac{4x^2y\sqrt{\frac{x^2+2y^2-1}{y^2}}+4x^3+4xy^2-4x}{y^4\sqrt{\frac{x^2+2y^2-1}{y^2}}}\right)$$
The minimum of the function will happen when $\nabla f(x,y) = 0$, thus we must solve for such a situation:
The first part:
$$\frac{4xy\sqrt{\frac{x^2+2y^2-1}{y^2}}+4x^2+4y^2-2}{y^3\sqrt{\frac{x^2+2y^2-1}{y^2}}}=0,$$ $$2xy\sqrt{\frac{x^2+2y^2-1}{y^2}}+2x^2+2y^2-1=0,$$ $$2x^2y\sqrt{\frac{x^2+2y^2-1}{y^2}}+2x^3+2xy^2-x=0$$
The second part:
$$\frac{4x^2y\sqrt{\frac{x^2+2y^2-1}{y^2}}+4x^3+4xy^2-4x}{y^4\sqrt{\frac{x^2+2y^2-1}{y^2}}} = 0,$$
$$x^2y\sqrt{\frac{x^2+2y^2-1}{y^2}}+x^3+xy^2-x = 0,$$ $$x^2y\sqrt{\frac{x^2+2y^2-1}{y^2}}=-x^3-xy^2+x$$
We can substitute the second part into the first part like so:
$$2(-x^3-xy^2+x)+2x^3+2xy^2-x=0$$ $$x=0$$
Thus, the minimum will occur when $x=0$. Thus, we plug in $x=0$ to the original function like so:
$$f(y)=\frac{1}{y^2}+2-\frac{1}{y^2}$$
where $x=0$, $-1 \le y \le0$, $y^2 \le 1$ and $2y^2 \ge 1$.
Which then reduces to:
$$f(y)=2$$
where $x=0$, $-1 \le y \le0$, $y^2 \le 1$ and $2y^2 \ge 1$.
However, the constraint $y^2 \le 1$ is equivalent to $-1\le y \le 1$, but is dominated by $-1 \le y \le 0$, and $2y^2 \ge 1$ is equivalent to $y^2 \ge \frac{1}{2}$ which is equivalent to $y \ge \frac{1}{\sqrt{2}}$ or $y \le -\frac{1}{\sqrt{2}}$.
Thus, the minimum will occur when $x=0$, and for any value of $y$ that satisfies $-1\le y \le -\frac{1}{\sqrt{2}}$.
I'd plot this in Desmos, see below:
https://www.desmos.com/calculator/d8fof2su7r
as you can see, the minimum lies tangent to the curve $x^2+y^2=1$. To find this point exactly, try to compute for which $x$ and $y$ these curves have equal gradient. Or substitute $x^2 = 1-y^2$ into $f$. This would give you a function with one variable, which you should be able to investigate in more detail.
