Use the inverse function theorem to find $y = G(u,v)$ given $u = f(x,y), v = g(x,y)$ Consider the functions $u = f(x,y) = x^2-y^2$ and $v = g(x,y) = 2xy$. What does the inverse function theorem tell you about defining $x = F(u,v)$ and $y = G(u,v)$? Find an expression for $\frac{\delta G}{\delta u}$.
I've struggled a bit with intuitively understanding the implicit function theorem, but I gather that if the Jacobian of the system of two equations has a non-zero determinant around a point $a = (x_0,y_0)$, then one can define the functions $F$ and $G$. I'm not sure how to explicitly find $\frac{\delta G}{\delta u}$, however. Any help appreciated!
 A: The functions $f$ and $g$ defined together a map $\varphi \colon \mathbb{R}^2 \rightarrow \mathbb{R}^2$ by $\varphi(x,y) = (u,v) = (x^2 - y^2, 2xy)$. The inverse function theorem tells you that if, at a point $(x_0,y_0)$, the linearization $d\varphi|_{(x_0,y_0)}$ of $\varphi$ is invertible, then the function $\varphi$ itself is locally invertible and differentiable. In our case, we have
$$ d\varphi|_{(x_0,y_0)} = \left. \begin{pmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{pmatrix} \right|_{(x_0,y_0)} = \begin{pmatrix} 2x_0 & -2y_0 \\ 2y_0 & 2x_0 \end{pmatrix} \implies \det \left( d\varphi|_{(x_0,y_0)} \right) = 4(x_0^2 + y_0^2).$$
Thus, $d\varphi|_{(x_0,y_0)}$ is invertible for all $(x_0,y_0) \neq (0,0)$ and so we can locally define $\psi(u,v) = (F(u,v),G(u,v))$ such that $\psi \circ \varphi = \operatorname{id}$ holds locally around such $(x_0,y_0)$. The chain rule then implies that
$$ d\psi|_{(u,v)} \circ d\varphi|_{(x,y)} = d(\operatorname{id})|_{(x,y)} = I_{2 \times 2} \implies \\
d\psi|_{(u,v)} = \left. \begin{pmatrix} \frac{\partial F}{\partial u} & \frac{\partial F}{\partial v} \\ \frac{\partial G}{\partial u} & \frac{\partial G}{\partial v} \end{pmatrix} \right|_{(u,v)} = \left( d\varphi|_{(x,y)} \right)^{-1} = \frac{1}{4(x^2 + y^2)} \begin{pmatrix} 2x & 2y \\ -2y & 2x \end{pmatrix}  $$ 
and so
$$ \frac{\partial G}{\partial u } = -\frac{y}{2(x^2 + y^2)}. $$ 
This solution uses the chain rule for functions $\mathbb{R}^2 \rightarrow \mathbb{R}^2$ (the functions $\varphi,\psi$) and gives you the full information about the differential $d\psi$ so you also get $\frac{\partial G}{\partial v}, \frac{\partial F}{\partial u}, \frac{\partial F}{\partial v}$. 
