In short: I'm trying to understand how dividing two partial derivatives gives a "non-partial" derivative.
Let me give the context, since there might be notational peculiarities specific to that domain.
Given the following system of ordinary differential equations: $$\frac{du}{dt} = f(u,v)$$
$$\frac{dv}{dt} = g(u,v)$$
and let $(u_0, v_0)$ be a steady solution: $f(u_0, v_0) = g(u_0, v_0) = 0$.
Murray (pp. 226-227) draws the picture of some possible reaction null clines for $f=0$, $g=0$ at a steady state $(u_0, v_0)$ (see picture below) and states that at $(u_0, v_0)$, the "gradient on $g=0$" fulfils
$$ \frac{dv}{du} \Bigg]_{g=0} = - \frac{\frac{\partial g}{\partial u}}{\frac{\partial g}{\partial v}} $$.
How does one get this identity? Why is there a negative sign?
This is a consequence of the implicit function theorem and implicit differentiation. If $g(u,v)=0$ has a local solution $v=\phi(u)$, then using the chain rule on $g(u,\phi(u))=0$ gives $$ 0=\partial_ug(u,\phi(u))+\partial_vg(u,\phi(u))·\phi'(u) $$ and solving for $\phi'(0)=\frac{dv}{du}$ gives the cited formula.