So, here's what I'm trying to prove:
Let $f$ be a bijective function defined on a subset of the real numbers, with inverse given by $\phi$. Suppose that $f$ is differentiable at $x \in Dom(f)$ so that $f'(x) \neq 0$. Suppose that $\phi$ is continuous at $y = f(x)$. Then, $\phi$ is differentiable at $y = f(x)$ and:
$$\phi'(y) = \frac{1}{f'(x)}$$
Proof Attempt:
The fact that $f$ is bijective guarantees the existence of the inverse function $\phi$. Now, consider the following limit:
$$\phi'(y) = \lim_{\Delta y \to 0} \frac{\phi(y+\Delta y)-\phi(y)}{\Delta y}$$
where $y = f(x)$ and $y + \Delta y = f(x+\Delta x)$. Since $f$ is injective, $\Delta y \neq 0$. Now, we can rewrite the limit above as:
$$\phi'(y) = \lim_{\Delta y \to 0} \frac{(x+ \Delta x)-x}{f(x+\Delta x)-f(x)}$$
$$\phi'(y) = \lim_{\Delta y \to 0} \frac{\Delta x}{f(x+\Delta x)-f(x)}$$
Now, we need to show that $\Delta y \to 0$ iff $\Delta x \to 0$. By the continuity of $f$ at $x$, which follows from its differentiability at $x$, we conclude that $\Delta x \to 0 \implies \Delta y \to 0$.
Suppose that $\Delta y \to 0$. Then:
$$\lim_{\Delta y \to 0} \Delta x = \lim_{\Delta y \to 0}[(x+\Delta x)-x] = \lim_{\Delta y \to 0} [\phi(y+\Delta y)-\phi(y)]$$
The limit above is 0 because $\phi$ is continuous at $y = f(x)$. So, it follows that $\Delta y \to 0 \implies \Delta x \to 0$. Using this, we re-write the expression for $\phi'(y)$:
$$\phi'(y) = \lim_{\Delta x \to 0} \frac{1}{\frac{f(x+\Delta x)-f(x)}{\Delta x}}$$
$$\phi'(y) = \frac{1}{f'(x)}$$
where the limit exists because of the limit law for quotients, which applies because $f'(x) \neq 0$. This proves that $\phi$ is differentiable at $y = f(x)$ and also proves that the formula for computing it holds.
Does the proof above work? If not, why? How can I fix it?