Let $x=f(t)$ and $y=g(t)$, where $f$ and $g$ are differentiable on an interval $[a,b]$. Then the slope of the line tangent to the curve at the point corresponding to $p\in [a,b]$ is $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}$$ at $p$, provided $f'(p)\neq 0$.
I was wondering if we also need $f'$ and $g'$ are continuous at $p$.
We shouldn't need any additional requirements, and that should be clear if we go through the reasoning behind this formula.
Suppose we have our curve parametrized by $\vec{r}(t) = f(t) \hat{e}_x + g(t)\hat{e}_y,$ where $f(t)$ and $g(t)$ are differentiable at $t_0.$
By the definition of a derivative,
$$\vec{r}'(t_0) = \lim_{t \to t_0} \frac{\vec{r}(t) - \vec{r}(t_0)}{t - t_0} = \lim_{t \to t_0} \frac{[f(t) \hat{e}_x + g(t)\hat{e}_y] - [f(t_0) \hat{e}_x + g(t_0)\hat{e}_y]}{t - t_0}$$
Now using the linearity of limits, we have
$$\vec{r}'(t_0) = \hat{e}_x \lim_{t \to t_0} \frac{f(t) - f(t_0)}{t - t_0} + \hat{e}_y \lim{t \to t_0} \frac{g(t) - g(t_0)}{t - t_0} = f'(t_0)\hat{e}_x + g'(t_0)\hat{e}_y$$
This vector points along the tangent to our graph, so if defined, the slope of the tangent line at $(f(t_0), g(t_0))$ must be $\frac{g'(t_0)}{f'(t_0)}.$ We are able to show this without the derivatives being continuous.