Let $f:\mathbb{R}^2 \to \mathbb{R}^2$ be a function given by $$f\left(x,y\right)=\left(x^2-y^2,2xy\right),\,\, \left(x,y\right) \in \mathbb{R}^2$$ Question : Compute $Df^{-1}\left(0,1\right)$.
I have computed that the solutions to $f\left(x,y\right)=\left(0,1\right)$ are : $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ and $\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)$. I know that by inverse function theorem I should get $Df^{-1}\left(0,1\right)=\left[Df\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\right]^{-1}$ or $\left[Df\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)\right]^{-1}$.
Problem : $(1)$ How do I check the conditions to apply inverse function theorem? In particular, what exactly do I need to check for this?
$(2)$ What guarantees us that $\left[Df\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\right]^{-1}$ and $\left[Df\left(-\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right)\right]^{-1}$ are same?
My intuition for $(1)$ is that it is enough to check that $f$ is differentiable at $\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)$ (since $Df$ is a linear map, it is automatically continuous and hence $f \in C^1$) and det$\left[Df\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)\right] \neq 0$. But I'm not sure if it is really enough. Any help would be much appreciated.