Consider the second order system
$$\dot{x}_{1}=\dfrac{-6x_1}{u^2}+2x_2 =f_1(x_1,x_2),~\dot{x}_{2}=\dfrac{-2(x_1+x_2)}{u^2}=f_2(x_1,x_2)$$
where $u=1+x^2_1$. Consider the hyperbola $x_2=\dfrac{2}{x_1-\sqrt{2}}$. Show, by investigating the vector field on the boundary of this hyperbola, that trajectories to the right of the branch in the first quadrant cannot cross that branch. I found this question in Exercise problem 4.8 of the book "Nonlinear Systems" by H. K. Khalil.
I first computed $\dfrac{f_2}{f_1}$ and $\dfrac{dx_2}{dx_1}$ at the hyperbola boundary $x_2=\dfrac{2}{x_1-\sqrt{2}}$.
$$\dfrac{f_2}{f_1}|_{x_2=\dfrac{2}{x_1-\sqrt{2}}}=\dfrac{-2x^2_{1}+2\sqrt{2}x_1-4}{2x^2_{1}+6\sqrt{2}x_1+4x^4_{1}+4}.$$
I was looking for some suggestions to approach this problem from here on. Any assistance in greatly appreciated.