Evaluating stability using Jacobian in dimension 1 Generally, when we are considering a system of differential equations, to evaluate the stability of fixed points, we find the eigenvalues of the Jacobian to evaluate stability. I was wondering how this applies to 1 dimensional systems. 
For example, when we have $\frac{dx}{dt}= -3x + \frac{5}{7+x}$, we can evaluate the stability of the fixed points by simply analyzing how it grows around the point (by plugging in points nearby). 
How would I use the Jacobian in such a system? 
In 2D systems, where $\frac{dx}{dt} = -3x + \frac{5}{7+y}$ and $\frac{dy}{dt} = -3y + \frac{5}{7+x}$, I would have to derive the respective equations with respect to $x$ and $y$ to form my Jacobian. Would I follow the same method but only with respect to $x$ for the 1D system? In that case, is the eigenvalue whatever value the derivative of $\frac{dx}{dt}$ with respect to $x$ is at a point?
 A: The Jacobian for scalar systems is itself a scalar.
We have $\dot{x}=f(x)\implies J = \left.\dfrac{df}{dx}\right|_{x=x_\text{eq}}$.
The equilibrium points of your system are given by  $$0=-3x_\text{eq}+\dfrac{5}{7+x_\text{eq}}\implies 3x_\text{eq}(7+x_\text{eq})=5\implies 3x^2_\text{eq}+21x_\text{eq}-5=0.$$
Solve this equation. Then determine the derivative of $f(x)=-3x+\dfrac{5}{7+x}$ and evaluate it at the equilibrium points. Your eigenvalue is given by the equation
$$\det\left[J-\lambda I \right]=0\implies \det[J-\lambda]=0\implies J-\lambda=0 \implies \lambda = J.$$
If the real part of your eigenvalue is positive the equilibrium point is unstable if the real part of your eigenvalue is negative the equilibrium is asymptotically stable. If the equilibrium point has a real part of $0$ we cannot conclude anything about the nonlinear system by Lyapunov's linearization theorem.

Just a remark: The differential equation is separable. Hence, you can determine the solution by integrating
$$\int_{\bar{x}=x_0}^{x}\dfrac{d\bar{x}}{-3\bar{x}+\dfrac{5}{7+\bar{x}}}=\int_{\tau=t_0}^td\tau$$
