Let say I have a 2D-dynamical system $$\dot x = f(x,y,\alpha_i)$$ $$\dot y = g(x,y,\alpha_i)$$ $i=1,2,...,n$ where $\alpha_i$ is a constant parameter.
Let $(x_0, y_0)$ be a fixed point, of which we want to determine its the stability.
Let us perturb the system such that $x = x_0 + \epsilon_x$ and $y = y_0 + \epsilon_y$.
Question:
If we know that when $\epsilon_x \not = 0$ and $\epsilon_y = 0$, $\epsilon_x \to 0$, and when $\epsilon_x = 0$ and $\epsilon_y \not = 0$, $\epsilon_y \to 0$, i.e the fixed point is stable along the each axis, can we conclude that the fixed point is stable?
No, we can't. The counterexample is $$ \dot r=-r^5 \cos (4\phi) \qquad \dot \phi= 0 $$ or, in Cartesian coordinates, $$ \dot x=-x^5+6x^3 y^2-xy^4\qquad \dot y=-x^4y+6x^2 y^3-y^5 $$ Notice that any ray $\phi= \text{const}$ is an invariant set (since $\dot\phi=0$). Let us find out the direction of motion along this ray depending on $\phi$. If the initial point is on the $x$- or $y$-axis, i.e. $\phi= \frac{\pi k}2,\;\;k\in\mathbb Z$, then $\cos (4\phi) = 1$ and $\dot r= -r^5$, so the direction is toward the origin. Сonsider the case $\phi= \pi/4$. In this case the direction is away from the origin: $\dot r= r^5$. Thus, the origin is unstable.