A choice of Lyapunov function for this 2D system? I am thinking of a  choice of a suitable Lyapunov function$V(x_{1},x_{2})$ which can make the system stable around the fixed point $x_{1}=1,x_{2}=1$
$\dot{x_{1}} = x_{1}x_{2} - x_{1}^2 $
$\dot{x_{2}} = x_{2} - x_{1}x_{2} +  2 -2x_{2}^2$
I thought of using $V(x_{1},x_{2}) = a(x_{1}-1)^2 + b(x_{2}-1)^2$, but then I am unable to show the $\frac{dV}{dt}  <0$ but still we have $V(1,1)  =0$, which I asked here - Determining $a$ and $b$ such that the expression is negative? and it seems I cannot take such a $V$? any help with the expression of $V(x_{1},x_{2})$?

You only need that $\dot V<0$ close to the sink $(1,1)$, so check the expression you originally tried, $V=a(x-1)^2+b(y-1)^2$ with
\begin{align}
\dot V&=2a(x−1)(xy−x^2)+2b(y−1)(y−xy+2−2y^2)\\
&=2a(x−1)x(y-x)+2b(y−1)(y(1-x)+2(1+y)(1-y)\\
&=-2ax(x-1)^2-2b(1+y)(y-1)^2~+~(2ax-2by)(x-1)(y-1)
\end{align}
Thus for $a=b=1$ you get for $(x,y)\approx(1,1)$ that $$\dot V\approx-2(x-1)^2-4(y-1)^2$$
which is negative. The remainder is of size $V^{3/2}$, thus smaller than the quadratic contributions close to $(1,1)$.
In some rough but exact estimate,
\begin{align}
\dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\\
&\le-2V+8V^{3/2}
\end{align}
which is negative for $4V^{1/2}<1$ or $V<\frac1{16}$. Using better estimates that combine some of the third order terms one might get larger bounds, 
\begin{align}
\dot V&=-2(x-1)^2-4(y-1)^2~-~2(x-1)^3-2(y-1)^3+2(x-1)^2(y-1)-2(x-1)(y-1)^2\\
&\le-2V+2\Bigl[(x-1)^2+(y-1)^2\Bigr]\bigl(|x-1|+|y-1|\bigr)\\
&\le-2V+2\sqrt2V^{3/2},
\end{align}
which means that the admissible region is $V<\frac12$.
