I'm trying to convince (prove) myself that if a set $S \subset \mathbb{({R^+})^2}$ is a convex region, then $S' := \left\{ \left(\frac1{x}, \frac1{y}\right) ; (x, y)\in S \right\}$ is also a convex region. I've tried using the definition of convexity, and the couple of convexity-preserving transformations (linear, affine, etc.) but I still can't establish my claim. Is my claim even true?
The closest I've gotten to show this is that $(x,y) \text{ is convex} \implies (x,y,1) \text{ is convex} \implies \left (1, \frac{y}{x}, \frac{1}{x}\right) \text{ is convex (perspective transform)} \implies \left (\frac1x, \frac{y}{x}\right) \text{ is convex}$
Here by $(x, y)$ I am abusing notation to refer to the set $S$.
PS: For the problem I'm actually trying to solve, $S$ is the interior of an ellipse, so even if my claim isn't true for general $S$, could it be true for this particular case?
It is not true, not even for an ellipse. A line segment (and hence, a sufficiently thin ellipse) may be warped by the transformation, see example picture below.