$f$ is a bijective function with differentiable inverse at a single point 
Let $\Omega \subseteq \mathbb{R}^n$ and $p \in \Omega$.  Let $f:U \to V$ be a bijection of open sets $p \in U \subseteq \Omega$ amd $f(p) \in V \subseteq \mathbb{R}^n$.  If $f^{-1}: V \to U$ is differentiable at $p$, then $df_p: \mathbb{R}^n \to \mathbb{R}^n$ is invertible.

Suppose $f$ is a bijection.  Since $f^{-1}$ is differentiable at $p$, it is continuous at $p$. Let $f^{-1}(f(x))=x$.  Then
\begin{align*}
df^{-1}(f(x))&=1\\
df_p^{-1}df_p&=1\\
\end{align*}
Following Dan Shved's suggestion I applied the chain rule, but I'm not sure that $f$ is differentiable - thus I'm not sure if $df_p$ exists.  If $f^{-1}$ were continuously differentiable this would be easier because I could invoke the Inverse Function theorem.

The problem statement is either incorrect, incomplete, or both. Certainly, in order to say anything about $df_p$, the assumption on $f^{-1}$ should be made at $f(p)$. But the mere fact that $f^{-1}$ is differentiable at $f(p)$ is not enough. For example, $f(x)=x^{1/3}$ is a bijection of $(-1,1)\subset \mathbb R$ onto itself. Its inverse $f^{-1}(x)=x^3$ is differentiable at $0=f(0)$, but $f$ is not differentiable at $0$. 
The following amended statement is correct:

Let $f:U \to V$ be a bijection of open sets $p \in U \subseteq \Omega$ amd $f(p) \in V \subseteq \mathbb{R}^n$.  If $f^{-1}: V \to U$ is differentiable at $\mathbf{f(p)}$, then $df_p: \mathbb{R}^n \to \mathbb{R}^n$ is invertible provided it exists.

Indeed, the chain rule applies to $\mathrm{id}=f^{-1}\circ f$ at $p$ and yields $\mathrm{id}=df^{-1}_{f(p)} \circ df_p$. Hence $df_p$ is invertible.
