For the existence, you may want to prove $f$ is monotone by examining $f^{\prime}$.
Find the range of $f$ (that will be the domain of $f^{-1}$) and examine where $f^{\prime}(x)\neq 0$. At the points $y=f(x)$ where $f^{\prime}(x)\neq 0$, $f^{-1}$ will be differentiable. In addition, $(f^{-1})^{\prime}(x)=\frac{1}{f^{\prime}(f^{-1}(x))}$
Proof of the last equality:
Since $f$ is continuous and 1-1 so without loss of generality we can suppose that $f$ is increasing. So for $c\in (a,b)=D_f$ so that $f^{\prime}(c)\neq 0$, \begin{equation}0\neq f^{\prime}(c)= \lim_{x\to c}\frac{f(x)-f(c)}{x-c}\Rightarrow \lim_{x\to c}\frac{x-c}{f(x)-f(c)}=\frac{1}{f^{\prime}(c)}\end{equation} Let $\epsilon>0$. Then, \begin{equation}\exists \delta_0>0: 0<\left|x-c\right|<\delta_0\Rightarrow \left|\frac{x-c}{f(x)-f(c)}-\frac{1}{f^{\prime}(c)}\right|<\epsilon\end{equation} In addition, $\exists \delta_1>0:[c-\delta_1,c+\delta_1]\subset (a,b)$. Let $\delta=\min{\left\{\delta_0,\delta_1\right\}}$
Let $y_1=f(c-\delta),y_2=f(c+\delta)$. Then since $f$ is increasing, \begin{equation}y_10:y_1$y\in (f(c)-\delta_2,f(c)+\delta_2)\subset (y_1,y_2)$ then by the Intermediate Value Theorem $\exists x\in (c-\delta,c+\delta):f(x)=y$. Thus, \begin{gather} 0<\left|y-f(c)\right|<\delta_2\Rightarrow \exists x\in (a,b):f(x)=y\text{ and } 0<\left|x-c\right|<\delta \Rightarrow \left|\frac{x-c}{f(x)-f(c)}-\frac{1}{f^{\prime}(c)}\right|<\epsilon\Rightarrow \notag\\ \left|\frac{f^{-1}(y)-f^{-1}(c)}{y-f(c)}-\frac{1}{f^{\prime}(c)}\right|<\epsilon\Rightarrow \lim_{y\to f(c)}\frac{f^{-1}(y)-f^{-1}(c)}{y-f(c)}=\frac{1}{f^{\prime}(c)} \end{gather} and so, $f^{-1}$ is differentiable at $f(c)$ and $(f^{-1})^{\prime}(f(c))=\frac{1}{f^{\prime}(c)}$