You are given the system $$\dot{x}=-x-xy^2; \dot{y}=2x^2y-x^2y^3$$
(a) What does the linearization about $x^*=(0,0)$ tell us about the local behavior. So $Df(x,y) = \begin{bmatrix} -1-y^2 & -2xy \\[0.3em] 4xy-2xy^3 & 2x^2-3x^2y^2 \end{bmatrix}$ So at $x^*=(0,0)$ we have the following $Df(0,0) = \begin{bmatrix} -1 & 0 \\[0.3em] 0 & 0 \end{bmatrix}$ with $\lambda_1=-1$, $\lambda_2=0$ Note that Re $\lambda_i < 0$ for $i=1,2$ Then $x^*$ is asymptotically stable
(b) Look for the Liapinouv function of the form $V(x,y)=ax^2+by^2$ for the equilibrium point $(0,0)$. Is it weak or strong Liapunov function? $$\dot{V}=2ax[-x-xy^2]+2by[2x^2y-x^2y^3]=-2ax^2-2ax^2y^2+4bx^2y^2-2bx^2y^4$$ I need help with part (b). I am not sure if the function is weak or strong
Hint for part (a): The origin is not asymptotically stable, so the right answer is that the linearization tells us nothing about the stability, although it certainly tells us something about the local behavior. Spoiler: take $x=0$ and see what you get.
Hint for part (b): Note that $$ -2ax^2-2ax^2y^2+4bx^2y^2-2bx^2y^4=-2x^2[a+(a-2b)y^2+by^4] $$ and so it cannot be a "strong" Lyapunov function (as you call it, the name is rather unusual), because it vanishes for $x=0$. Spoiler for the rest: for $(x,y)$ very small, $a+(a-2b)y^2+by^4$ is approximately $a\ne0$, and so it is a "weak" Lyapunov function.