If I have $g$ to be the inverse function of a continuous, strictly rising function $f:I\rightarrow\mathbb{R}$ and $f$ is differentiable in $f'(y_0)\ne 0$, then g is differentiable in $x_0=f(y_0)$ with
$$g'(x_0)=\frac{1}{f'(y_0)}=\frac{1}{f'(g(x_0))}.$$
Proof:
There is a function $\varphi:I\rightarrow\mathbb{R}$ continuous in $y_0$ s.t.
$$f(y)-f(y_0)=\varphi(y)\cdot(y-y_0)$$
and $\varphi(y_0)=f'(y_0)$. Because $f$ is strictly rising and $f'(y_0)\neq 0,\varphi(y)\neq 0$ for all $y\in I$, with $x=f(y),y=g(x)$ one gets
$$g(x)-g(x_0)=\frac{1}{\varphi(g(x))}\cdot(x-x_0).$$
Because $\frac{1}{\varphi\circ g}$ is continuous, $g$ is differentiable in $x_0$ with $\frac{1}{f'(y_0)}$.
My question is why is $\frac{1}{\varphi\circ g}$ continuous and how do we know that $\frac{1}{\varphi(y_0)}=\frac{1}{f'(y_0)}?$