Find the minimum possible value of $x+y-2xy$ over reals. Find the minimum possible  value of $x+y-2xy$, if $x+y\ge 3xy,~ 2(x+y)\ge 1+3xy~$ and $~0\le x\le y\le 1.$

My guess is $x=y=\frac{2}{3},$ gives the minimum value of $\frac{4}{9}$.
I thought this manipulation might be useful,
$$x+y-2xy=\frac{3(x+y)-6xy}{3}\ge \frac{1}{3}$$
But, $x+y-2xy=\frac{1}{3}$ would imply $x+y=3xy$ and $2(x+y)=1+3xy\implies x+y=1$ and $xy=\frac{1}{3}.$ This system does not have real solutions and hence $\frac{1}{3}$ is not an attainable value.
Another idea was to substitute $0\le x+y=s\le 2~$ and $~0\le xy=p\le 1$, transforming the problem into finding $\min(s-2p)$ given the conditions,
$$s\ge 3p, ~ 2s\ge 1+3p, ~ s^2\ge 4p.$$

I think that your second idea works.
Let $k:=s-2p$, and $f : p=\frac 14s^2, g : p=\frac 23s-\frac 13$ and $h : p=\frac s3$.
Then, $f$ and $g$ intersect at $A(2/3,1/9)$ and $C(2,1)$, and $f$ and $h$ intersect at $O(0,0)$ and $B(4/3,4/9)$.
So, one can see that

*

*for $s\lt\frac 23$, $p\leqslant \frac 23s-\frac 13$


*for $\frac 23\leqslant s\leqslant \frac 43$, $p\leqslant \frac 14s^2$


*for $\frac 43\lt s$, $p\leqslant \frac s3$
Since the slope of the line $k=s-2p$, i.e. $p=\frac 12s-\frac k2$ is $\frac 12$, the minimum possible value of $k$, i.e. the largest $p$-intercept is attained when the line passes through $A$ and $B$.
So, the minimum possible value of $k$ is $k=s-2p=\frac 23-2\times \frac 19=\color{red}{\frac 49}$ which is attained when $(s,p)=(2/3,1/9),(4/3,4/9)$, i.e. when $(x,y)=(1/3,1/3),(2/3,2/3)$.
