Compute $Df^{-1}(2,5)$ given $f(u,v)=(uv,u^2+v^2)$ using the Inverse Function Theorem (IFT) I need help figuring out where my solution below goes wrong, because the question was phrased as if the derivative would exist.
$$f(u,v)=(uv,u^2+v^2)$$
We prove $D {f}^{-1}(2,5)$ does not exist. Clearly $ {f}$ is $\mathcal{C}^1$ on $\mathbb{R}^2$. Note that $ {f}(1,2)=(2,5)= {f}(2,1)$. We consider the Jacobian of $ {f}$ at $(1,2)$ and at $(2,1)$:
$$\Delta_{ {f}}(1,2)=\begin{vmatrix} 2 & 1 \\ 2 & 4 \end{vmatrix}=2(4-1)=6.$$
$$\Delta_{ {f}}(2,1)=\begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix}=2(1-4)=-6.$$
Since $\Delta_{ {f}}(1,2)\neq 0$ and $\Delta_{ {f}}(2,1)\neq 0$, by the IFT there exists an open $V$ and $W$ respectively in $\mathbb{R}^2$ such that $ {f} ^{-1} $ is $\mathcal{C}^1$ on $ {f}(V)$ and $ {f}(W)$. We know $ {f}$ is $\mathcal{C}^1$ on $\mathbb{R}^2$ and $V$, $W$ are open in $\mathbb{R}^2$, thus $ {f}(V)$, $ {f}(W)$ are open in $\mathbb{R}^2$. Note that $ {f}( {f} ^{-1}(2,5))= {f}(1,2)= {f}(2,1)$. Since $(2,5)= {f}(1,2)\in {f}(V)$ and $(2,5)= {f}(2,1)\in {f}(W)$ by the IFT we have:
$$D {f} ^{-1}(2,5)  =  [D {f}( {f} ^{-1}(2,5))] ^{-1}  =  [D {f}(1,2)] ^{-1} $$
and
$$D {f} ^{-1}(2,5)  =  [D {f}( {f} ^{-1}(2,5))] ^{-1}  =  [D {f}(2,1)] ^{-1}.$$
Thus $[D {f}(1,2)] ^{-1} =[D {f}(2,1)] ^{-1} $. We know $[D {f}(u,v)] ^{-1} =\begin{bmatrix} v & u \\ 2u & 2v \end{bmatrix} ^{-1} $ so for $|u|\neq|v|$ we have:
$$\displaystyle[D {f}(u,v)] ^{-1} =\frac{1}{2v^2-2u^2}\cdot\begin{bmatrix} 2v & -u \\ -2u & v \end{bmatrix}.$$ Since $1\neq 2$ we have:
$$D {f} ^{-1}(2,5)=[D {f}(1,2)] ^{-1} =\frac{1}{6}\cdot\begin{bmatrix}
    4 & -1 \\ -2 & 2
\end{bmatrix}$$
$$D {f} ^{-1}(2,5)=[D {f}(2,1)] ^{-1} =-\frac{1}{6}\cdot\begin{bmatrix}
    2 & -2 \\ -4 & 1
\end{bmatrix}$$
However, it is clear that $\displaystyle \frac{1}{6}\cdot\begin{bmatrix}
    4 & -1 \\ -2 & 2
\end{bmatrix}\neq -\frac{1}{6}\cdot\begin{bmatrix}
    2 & -2 \\ -4 & 1
\end{bmatrix}$. Thus $D {f} ^{-1}(2,5)\neq D {f} ^{-1}(2,5)$ a contradiction, so $D {f} ^{-1}(2,5)$ does not exist.
What I really don’t understand is that this contradiction implies that $Df^{-1}(2,5)$ does not exist, thus $f^{-1}$ is not differentiable. This would contradict the IFT since one of the implications of the theorem is that $f^{-1}$ must be differentiable on the open intervals $f(V)$/$f(W)$.

Notice that you're applying the IVT twice here. Once at the point $(1,2)$ and another times around $(2,1).$
By applying the IFT around the point $(1,2)$ you get an open set $W_1$ containing $(1,2)$ and a second open set $V_1 = f(W_1)$ containing $f(1,2) = (2,5)$. The IFT tells you that the restriction of $f$ to $W_1$ is invertible and that it's inverse $f_{W_1}^{-1}$ is also differentiable.
By applying this theorem again at the point $(2,1)$ you will find open sets $W_2$ and $V_2 = f(W_2)$ and the IVF tells you that these the restriction of $f$ to $W_2$ , $f_{W_2}$ is invertible and that its inverse $f_{W_2}$ and that it's inverse $f_{W_2}^{-1}$ is also differentiable.
We are therefore dealing with two different functions here : $f_{W_1}$ and $f_{W_2}$ !
In your solution you do not take this into account and are giving these functions the same name $f$. The apparent contradiction that you found isn't one because you're calling two different matrices $D{f_{W_1}^{-1}}(2,5)$ and $Df_{W_2}^{-1}(2,5)$ by the same name $Df^{-1}(2,5)$.
It might be surprising to you that
$$ D{f_{W_1}^{-1}}(2,5) \neq Df_{W_2}^{-1}(2,5)$$
but when you think about it $f_{W_1}$ and $f_{W_2}$ are just different functions so one shouldn't always expect the above quantities to be same. That doesn't mean that they aren't related however !
I suspect that we have $f_{W_1} = f_{W_2} \circ \pi_{W_1}$ where $\pi_{W_1}$ is  the smooth invertible function
$$ \pi_{W_1} : W_1 \rightarrow W_2 : x,y \mapsto \pi_{W_1}(x,y) = (y,x)$$
We would then have $f_{W_1}^{-1} = \pi_{W_1}^{-1}\circ f_{W_2}^{-1}$ and using the chain rule you will find that
$$ Df_{W_1}^{-1}(2,5) = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \times Df_{W_2}^{-1}(2,5).$$
