What does the lack of singular points of non liniar system mean in phase portraits? I'm going to plot the phase portrait of this system: $\dfrac{dx}{dt}=-x^2 + 4 y^2$ $\dfrac{dy}{dt}=-8 - 4 y + 2 x y$ The singular point $(x,y)$ can be found from the system: $-x^2 + 4 y^2=0$ $-8 - 4 y + 2 x y=0$ Seems this system doesn't have roots wolframalfa. What does it mean? I have integral curves(phase trajectories) that do not intersect? If yes, hence, obviously I can't determine type of the singular point ? By the way, I heard from my lector, that it can be drawn in Wolframalfa. Can you provide an example respectively to my system?

The system that you have given has indeed two singular points: $$x_1 = -2,\qquad y_1 = -1,$$ and $$X_2 = 4, \qquad y_2 = 2;$$ You can verify this by checking that $dx/dt$ and $dy/dt$ are both zero.

To check what type of singular points these are, we can study the system linearized around a stable point: $$ \dot x \approx A x $$ where $A = \
abla f (\bar x)$. In this case, we get that $$ A(x,f) = \
abla f(x,y) = \begin{bmatrix} -2x & 8y \\\ 2y & -4+2x \end{bmatrix}.$$

This matrix has two eigenvalues, $$ \lambda_{1,2}(x,y) = -2 \pm \sqrt{4-8x + 4x^2 + 16x^2},$$ of which, in the case of the two given stable points, one is positive: $$ \lambda_1(x_1,y_1) \approx 5.21$$ $$ \lambda_1(x_2,y_2) = 8 $$ Because the linearized system has positive eigenvalues, the singular points are unstable.

To draw the phase portrait you can use a vector field plot).