Given a general ellipsoid equation $x^\top Q x \leq 1$, where $x\in\mathbb{R}^d$ and $Q\in\mathbb{S}_+^d$ is positive semidefinite. It is clear that the feasible region of $x$ corresponds to an ellipsoid hence it is convex.
Now if we define $y_i = x_i^2,~\forall i\in [d]$. Is the following new feasible region $\mathcal{F}$ of $y$ still convex?
$$\mathcal{F}:= \{y\mid y_i = x_i^2, x_i \geq 0, x^\top Q x \leq 1\}$$
Here is an idea how to construct a counterexample in $d=3$: Define $a:=(1/\sqrt2, 1/\sqrt2,0)$ and $b=(1/\sqrt2, 0, 1/\sqrt2)$.
In order that $F$ is convex, the point $ c:= (1/\sqrt2,1/2,1/2)$ needs to be in the ellipsoid, as it is the component-wise square-root of the mean of the component-wise squares of $a,b$.
However, the points $a$ and $b$ belong to the plane $\{x: \ x_1-x_2-x_3=0\}$, while $c$ does not belong to that plane. Now it should be possible to construct an ellipsoid that contains $a,b$ but not $c$.