How to study the behavior of points in the neighborhood of critical points 
Given the predator-prey system:
$$\frac{dx}{dt} = 12x -2x^2 - xy$$
$$\frac{dy}{dt} = 8y -4y^2 + 2xy$$
Where $f(x,y) = 12x -2x^2 - xy$, $g(x,y) = 8y -4y^2 + 2xy$
a) How will the point $(3,0)$ behave over time? Will the point converge to or diverge from the critical points?
b) How will the point $(3,1)$ behave over time? Will the point converge to or diverge from the critical points?

NOTE: I've written about how to get the critical points, linearize the system and solving the eigenvalue problem so that the problem is contextualized. However my issue is on how to analyse the behavior of points in the neighborhood  of the critical points. 

Let me explain what I've done so far:
Firstly I've found the critical points of the system (i.e. for what points $f(x,y)=0$ and $g(x,y)=0$ hold) : 
$$C_1 (0,0)$$
$$C_2 (0,2)$$
$$C_3 (6,0)$$
$$C_4 (4,4)$$
Secondly I've linearized the system $(2 \times 2)$ Jacobian matrix (first row derivatives of $f(x,y)$ with respect to x and y respectively; second row derivatives of $f(x,y)$ with respect to x and y respectively):
$$J=
    \begin{pmatrix}
    12 -4x -y & -x  \\
    2y & 8-8y+2x  \\
    \end{pmatrix}
$$
Thirdly I've solved the eigenvalue problem for each point and determined its 'nature':
1) For $C_1 (0,0)$ we get $\lambda_1 = 12$ and $\lambda_2 = 8$. As both eigenvalues are real, unequal and have the same (positive) sign the critical point $C_1 (0,0)$ is a nodal source; it is unstable.
2) For $C_2 (0,2)$ we get $\lambda_1 = 10$ and $\lambda_2 = -8$. As both eigenvalues are real, unequal and have opposite sign the critical point $C_2 (0,2)$ is a saddle point; it is unstable.
3) For $C_3 (6,0)$ we get $\lambda_1 = 20$ and $\lambda_2 = -12$. As both eigenvalues are real, unequal and have opposite sign the critical point $C_3 (6,0)$ is a saddle point; it is unstable.
4) For $C_4 (4,4)$ we get $\lambda_1 = -12 + 4i$ and $\lambda_2 = -12 - 4i$. As both eigenvalues are (negative) complex numbers with nonzero real part the critical point $C_4 (4,4)$ is a spiral sink; it is asymptotically stable.

OK this is fine but now it comes my problem. I do not really know how to proceed when given points in the neighborhood of the critical points and being asked to explain whether they depart or approach the critical points.
I tried plugging in the points to see how the system behaves. 
a) On the one hand we see that for $(3,0)$ the predator population does not change because $\frac{dy}{dt} = 0$. On the other hand, the prey population does change $\frac{dy}{dt} = 18$ (my intuition tells me that as we do not get $3$, the prey population must change).
Thus we already know that it will depart from the critical points $C_2 (0,2)$ and $C_2 (4,4)$ because its $y$ value isn't zero.
Let's now study the other two critical points:
Since $C_1 (0,0)$ is a nodal source, $(3,0)$ will never approach that critical point. 
As $C_4 (6,0)$ is a spiral sink and its y value is zero, $(3,0)$ will end up approaching $C_4$ critical point.
b) For $(3,1)$ Here you do not get neither $\frac{dy}{dt} = 0$ nor $\frac{dx}{dt} = 0$ I was quite stuck here so I had a look at the solution:
'For the point $(3,1)$ note that the solutions passing through the x axis or the y axis stay on those axis. This means that a solution with $x(0) \ge 0$ and $y(0) \ge 0$ will satisfy $x(t) \ge 0$ and $y(t) \ge 0$ for all $t \ge 0$. In this case we get $x(0) = 3$ and $y(0) = 1$, which means:
$$\frac{dx}{dt} = 12x - 2x^2 - xy \le 12x - 2x^2$$ 
And that means that $x(t)$ cannot run to infinity if $t \rightarrow \infty$. Thus there must be a $C > 0$ such that $x(t) \le C$ for all $t \ge 0$. Then we have that:
$$\frac{dy}{dt} = 8y - 4y^2 + 2xy \le (8+2C)y - y^2$$ 
And we analogously argue that $y$ must also be limited.
We conclude that $(x(t), y(t))$ must converge to either $C_2 (0,2)$, $C_3 (6,0)$ or $C_4 (4,4)$. As the point approaches a saddle point only if the solution is solely on the x axis or the y axis, we conclude that $(x(t), y(t))$ only ends up approaching the critical point $C_4 (4,4)$.'
I do not understand the above explanation (more precisely, I do not see why the solutions passing through the x axis or the y axis stay on those axis). 
How to deal with points in the neighborhood of critical points then? 
 A: *

*$C_3(6,0)$ is a saddle point, only $C_4(4,4)$ is a sink.

*You also need to consider the eigenvectors of the saddle points on the axes to see if any solution can converge towards them from the inside of the quadrants.



Then to the axes:


*

*The curves $(x(t),0)$ where $x$ is a solution of $\dot x=xf(x,0)$ and on the other axis $(0,y(t))$ where $y$ solves $\dot y=yg(0,y)$ are solutions of the given system that fill the coordinate axes completely. By the uniqueness theorem---the right sides are polynomial, thus smooth---any IVP with IC on the coordinate axes stays on that axis. No solution with IC not on the axes can have points on the axes or cross them.

*a) With the IC $(3,0)$ the dynamic reduces to one dimension, $\dot x=x(12-2x)$. On the interval $(0,6)$, the forward motion is to the right towards $C_3(6,0)$.

*b) The IC $(3,1)$ is inside the first quadrant, thus the solution stays inside the first quadrant. By the first inequality $$\dot x\le 2x(6-x)$$ the line $x=6$ can only be passed towards the left, decreasing $x$. No solution starting left can thus reach values greater than $C=6$. Then in the second equation $$\dot y\le2y(4+6-2y),$$ by the same argument no solution starting below $y=5$ (and $x\le 6$) can grow above $y=5$. A solution that stays inside a compact set has to converge towards some limit object, as $C_4(4,4)$ is a sink and not a source, there is no limit cycle, all solutions with IC in $[0,6]\times[0,5]$ converge to this sink.

