How to find slope of locus? Given a system of DEs: $$\begin{cases} \dot{x} = g_1(x,y) \\ \dot{y} = g_2(x,y), \end{cases}$$
where $g_1,g_2 \in C^1$
How to show that the slope of $g_1 = 0$ in the neighborhood of the steady state $A$ is equal to the negative fraction of the first column entry to the second column entry of the first row of $J(A)$, where $J$ is the Jacobian matrix? Can you give a hint?
 A: Let $(x_0,y_0)$ be some fixed point of your system of DEs, i.e.
$$g_1(x_0,y_0) = 0\quad\text{and}\quad g_2(x_0,y_0) = 0$$
Consider the level set $L_0 = \{(a,b)\in \mathbb R^2: g_1(a,b)=0\}$, which obviously contains $(x_0,y_0).$
Assuming in particular that
$$\frac{\partial g_1}{\partial x} (x_0,y_0) \neq 0,$$
then the piece of the level set $L_0$ which contains $(x_0,y_0)$ is (locally) a curve passing through $(x_0,y_0)$ (this is the Implicit Function Theorem) and so it makes sense to talk about a derivative—indeed, for your fraction to make sense you need this assumption as well.

So this is a cartoon of what the situation looks like locally (i.e. if you zoom in on $(x_0,y_0)$): namely, it looks like the graph of a function $f$ in $x$, and you are asking for the gradient at the point $(x_0,y_0)$, i.e. $f'(x_0)$. This corresponds to the gradient triangle in the picture
Now I'll sketch the details of this, which gives you the formula you want. 
Assuming that
$$\frac{\partial g_1}{\partial x} (x_0,y_0) \neq 0,$$
the Implicit Function Theorem says you can parametrise the curve which passes through $(x_0,y_0)$—at least some small piece of it. I.e., you can parametrise this piece of curve in $L_0$ by
$$t \mapsto (t,f(t)) \qquad(t \in (x_0-\delta,x_0+\delta)),$$
for some $\delta > 0$, and where $f is C^1$ and satisfies $f(x_0) = y_0$. I.e. for all $|t|< \delta$ we have
$$ g_1(t,f(t)) = 0$$
As mentioned above, the gradient you want is given by $f'(x_0)$. Differentiating this last formula gives, by the chain rule for partial derivatives,
$$ \frac{\partial g_1}{\partial x}(t,f(t)) + f'(t)\frac{\partial g_1}{\partial y}(t,f(t)) = 0;$$
and setting $t = x_0$ then gives
$$ \frac{\partial g_1}{\partial x}(x_0,y_0) + f'(x_0)\frac{\partial g_1}{\partial y}(x_0,y_0) = 0,$$
which rearranges to exactly what you want for $f'(x_0)$:
$$f'(x_0) = -\frac{\partial g_1}{\partial x}(x_0,y_0)\bigg/\frac{\partial g_1}{\partial y}(x_0,y_0).$$
Again, the left hand side is the gradient of the level curve through $(x_0,y_0)$. You don't actually need any information about $g_2$.
