Using Lyapunov function, prove that a critical poin (0,0) is asymptotically stable. Let a linear system
$$
\ x'=-2x+x{y}^{2}\\
\ y'=-x^{3}-y\\
$$
Using Lyapunov function, prove that  a critical point (0,0) is asymptotically stable.

The most obvious try for a Lyapunov function designed to show local stability of a differential system at $(0,0)$ might be 
$$
H(x,y)=x^2+y^2.
$$
In the present case,
$$
\frac{\mathrm d}{\mathrm dt}H(x,y)=-4x^2-2y^2+2xy(xy-x^2).
$$
Using the bounds $|xy-x^2|\leqslant|xy|+x^2$ and $2|xy|\leqslant x^2+y^2$, one sees that
$$
\frac{\mathrm d}{\mathrm dt}H(x,y)\leqslant-4x^2-2y^2+(x^2+y^2)\tfrac12(3x^2+y^2).
$$
Let $D=\{(x,y)\mid x^2+y^2\leqslant2\}$. For every $(x,y)$ in $D$,
$$
\frac{\mathrm d}{\mathrm dt}H(x,y)\leqslant-4x^2-2y^2+(3x^2+y^2)=-(x^2+y^2),
$$
that is,
$$
\frac{\mathrm d}{\mathrm dt}H(x,y)\leqslant-H(x,y).
$$
Hence, if $(x(0),y(0))$ is in $D$, then $(x(t),y(t))$ stays in $D$ for every $t\geqslant0$ and, solving the differential inequality $u'\leqslant-u$, one sees that 
$$
H(x(t),y(t))\leqslant H(x(0),y(0))\mathrm e^{-t}.
$$ 
In particular $H(x(t),y(t))\to0$ hence $(x(t),y(t))\to(0,0)$ when $t\to\infty$.
