Find a critical point I found a method to find a critical control parameter in one article and it worked for other similiar system with saddle-node bifurcation too. In general, we have a 2 equations nonlinear system:
$$\begin{align} 
&\dot{x}=f_1(x,y) + \mu \\
&\dot{y}=f_2(x,y)  
\end{align}$$
First of all we find nullclines $\dot{x}=0$ and $\dot{y}=0$ and to find critical parameter in which bifurcation happens $\mu = \mu_{c}$ enough to solve $\partial\dot{x}/\partial x = 0$, from which we can find critical $x^*$ and after that from nullclines value of $\mu_c$. Can somebody explain why $\partial\dot{x}/\partial x = 0$ works here? Is it somehow related with tangents or what? 
 A: This is not correct. Suppose that $f_1(x,y)+\mu = 0$ can be represented for a choose $\mu$ value, as the red curve in the following plot. Suppose also that $f_2(x,y)=0$ is also represented in blue. Those curves can intersect one three or be tangent. The tangency determination can be obtained by solving for $x,y,\lambda$ 
$$
\cases{
\vec n_1 = \lambda \vec n_2\\
f_2(x,y) = 0
}\ \ \ \ \ \ \ (1)
$$
where $\vec n_1 = \nabla f_1,\ \ \vec n_2 = \nabla f_2$ Generally speaking $\vec n_1$ have both components non null. Now focusing a case study
$$
\dot x = 2+3x-x^3-y+\mu\\
\dot y = \frac{17.1-3y(1+e^{-\tau x})}{2+e^{-\tau x}}
$$
for $\tau = 3.3$ we have for $f_1, f_2$ the graphics.

and the tangency point in black is obtained by solving $(1)$ obtaining $x_c = 0.6965, y_c = 5.17981,\lambda = 0.631873, \mu_c = 1.42819$ with $\vec n_1 = (1.54466,-1), \vec n_2 = (2.45156, -1.57171)$ as we can observe in the following plot.

Note that using the condition $\frac{\partial\dot x}{\partial x}=0$ the point found would be at the intersection with the vertical dashed green line.
Now if instead $\tau = 100$ we will find

and the tangency at $x_c = 1, y_c = 5.7, \lambda = 0.6666, \mu_c = 1.7$

Note that in this last case, as $e^{-100 x}\approx 0$ for $x > 0$ we have $\vec n_1 = (0,-1), \vec n_2 = (0,-1.5)$ 
