Suppose $x=f(s,t)$ and $y=g(s,t)$, what is $\frac{dy}{dx}$? Suppose $x=f(s,t)$ and $y=g(s,t)$, how do we calculate 
$$\frac{dy}{dx}$$
Should I use chain rule? I looked at total derivative, I got 
$$\frac{\text{d}y}{\text{d}x}=\frac{\partial g}{\partial s}\frac{\text{d}s}{\text{d} x}+\frac{\partial g}{\partial t}\frac{\text{d}t}{\text{d} x}$$
 A: To expand on my comment, we start with equations $x = f(s, t), y = g(s, t)$. The derivative $dy/dx$ makes sense only when $x, y$ are dependent variables i.e. there is some relation of the form $\phi(x, y) = 0$ between them. Differentiating this relation with respect to $s$ and $t$ we get $$\frac{\partial \phi}{\partial x}\frac{\partial x}{\partial s} + \frac{\partial \phi}{\partial y}\frac{\partial y}{\partial s} = 0, \frac{\partial \phi}{\partial x}\frac{\partial x}{\partial t} + \frac{\partial \phi}{\partial y}\frac{\partial y}{\partial t} = 0$$ And eliminating the derivaives of $\phi$ we get the Jacobian $$\frac{\partial(x, y)}{\partial(s, t)} = \begin{vmatrix}\dfrac{\partial x}{\partial s} & \dfrac{\partial x}{\partial t}\\ \dfrac{\partial y}{\partial s} & \dfrac{\partial y}{\partial t}\end{vmatrix} = 0$$ Note that the above condition is also sufficient for existence of a functional equation between $x, y$ and then the derivative $dy/dx$ is given by the relation $d\phi = 0$ or $$\frac{\partial \phi}{\partial x}\cdot dx + \frac{\partial \phi}{\partial y}\cdot dy = 0$$ or $$\frac{dy}{dx} = -\frac{\partial \phi/\partial x}{\partial \phi/\partial y} = \frac{\partial y/\partial s}{\partial x/\partial s} = \frac{\partial y/\partial t}{\partial x/\partial t}$$ or $$\frac{dy}{dx} = \dfrac{\dfrac{\partial g}{\partial s}}{\dfrac{\partial f}{\partial s}} = \dfrac{\dfrac{\partial g}{\partial t}}{\dfrac{\partial f}{\partial t}}$$ As a simple example let $$x = f(s, t) = s + t, y = g(s, t) = s + t$$ so that $y = x$ and $dy/dx = 1$. Clearly we have $\partial y/\partial s = 1, \partial x/\partial s = 1$ so that $dy/dx = \dfrac{\partial y/\partial s}{\partial x/\partial s} = 1$.
To convince yourself you can try a slightly complicated example $$x = f(s, t) = s + t, y = g(s, t) = (s + t)^{2} = s^{2} + t^{2} + 2st$$
