Why is it true that if the inverse of both $ \tilde{f} $ and $ f $ exists then:
$$ \tilde{f}\left(\vec{x}\right) = [Df(x_{0})]^{-1} f(\vec{x}) $$
$$ \implies \tilde{f}^{-1}(\vec{x}) = Df(\vec{x}_{0}) \space \space f^{-1}(\vec{x}) $$
Where $ Df(\vec{x}_{0}) $ is the jacobian of $ f $ evaluated at a particular point $ \vec{x}_{0} $. And $ [Df(\vec{x}_{0})]^{-1} $ denotes its inverse (assuming it exists as well). Also assume that the derivative of $ $ $ f $ at $\vec{x_{0}} $ exists. And that the domain and images of $ f $ and $ \tilde{f} $ are open.
Let us work in one dimension and pick $f(x)=(x+1)^2$ as an example, with $x_0=1$. Restrict the domain of $f$ to $x>-1$ in order to make it invertible. Then $Df(0)=2$, so $\tilde f(x)=\frac12(x+1)^2$.
Now $f^{-1}(y)=-1+\sqrt{y}$ and $\tilde f^{-1}(y)=-1+\sqrt{2y}$. So the claim reduces to $$-1+\sqrt{2y}=2(-1+\sqrt{y}).$$ Is this true for all $y>0$? I doubt it!
So what is true? Returning to the general case (but dropping the arrows on the vectors), one finds that $\tilde f(x)=y$ is equivalent to $f(x)=Df(x_0)y$, so $x=f^{-1}(Df(x_0)y)$. In other words, $$\tilde f^{-1}(y)=f^{-1}\bigl(Df(x_0)y\bigr),$$ which is a rather different formula than the one you have been given.