We present two proofs here, a short informal version and a longer, more formal version.

Short Informal Proof
Given \( y(t) \) and \( x(t)\), we can write \( y = y(x(t)) \). We want to find \( dy/dx \).
Using the chain rule on \( y(x(t)) \), we have \( dy/dt = dy/dx \cdot dx/dt\).
Solving for \( dy/dx \) we have \( \displaystyle{ dy/dx = \frac{dy/dt}{dx/dt} }\)      [qed]

Longer More Formal Proof
Given the two points \((x_1,y_1) = (X(t), Y(t))\) and \((x_2,y_2) = (X(t+\Delta t), Y(t+\Delta t)) \) and \(\Delta t > 0 \) on a smooth curve, let \( \Delta x = x_2 - x_1 = X(t+\Delta t) - X(t) \) and \( \Delta y = y_2 - y_1 = Y(t+\Delta t) - Y(t) \).

As \( \Delta t \to 0 \), we know that \(\Delta x \to 0 \) and we can write

\(\displaystyle{ \frac{dy}{dx} = \lim_{\Delta x \to 0}{\frac{\Delta y}{\Delta x}} = \lim_{\Delta t \to 0}{\frac{Y(t+\Delta t) - Y(t)}{X(t+\Delta t) - X(t)}} }\)

Now we can multiply the numerator and denominator by \(1/\Delta t\) and use limit laws to give us

\( \begin{array}{rcl} \displaystyle{\frac{dy}{dx}} & = & \displaystyle{\lim_{\Delta t \to 0}{\frac{[Y(t+\Delta t) - Y(t)](1/\Delta t)}{[X(t+\Delta t) - X(t)](1/\Delta t)}} } \\ & = & \displaystyle{ \frac{\lim_{\Delta t \to 0}{[Y(t+\Delta t) - Y(t)]/\Delta t}}{\lim_{\Delta t \to 0}{[X(t+\Delta t) - X(t)]/\Delta t}} } \\ & = & \displaystyle{ \frac{dy/dt}{dx/dt} ~~~~~ \text{ [qed] } } \end{array} \)