$x,y$ are functions of $u,v$ and $x=2uv, y=u^2-v^2$. There's an inverse map from such that $u,v$ are functions of $x,y$. Prove that $$||\nabla u||^2+||\nabla v||^2=\frac{0.5}{\sqrt{x^2+y^2}}$$
This is an exercise for implicit differentiation. I thought of the following: $$ u_u=v_x 2v+v_y 2v\\ v_v=v_x 2u-v_y 2v\\ u_u=u_x 2v+u_y 2u\\ u_v=u_x 2u-u_y 2v $$ Even if this is correct I don't see how we can factor out the gradients from here.
Differentiate $x=2uv$ and $y=u^2 - v^2$ each with respect to $x$ and with respect to $y$, to get four equations:
$$\begin{cases} u_x v + uv_x = 0.5 \ \\ uu_y - vv_y = 0.5 \\ u_yv + uv_y = 0 \\ uu_x - vv_x = 0 \end{cases}$$
Multiply the last equation by $v$ to get: $(vu_x)u = v^2 v_x$, and substitute $vu_x = 0.5 - uv_x$ to get:
$$v_x = \frac{u}{2(u^2 + v^2)}$$
which gives (using the last equation once more):
$$u_x = \frac{v}{2(u^2+v^2)}$$
Doing the same thing to the other equations we get:
$$v_y = -\frac{v}{2(u^2 + v^2)}$$
and:
$$u_y = \frac{u}{2(u^2+v^2)}$$
It's then clear that:
$$\|\nabla u\|^2 + \|\nabla v\|^2 = \frac{0.5}{(u^2 + v^2)}$$
notice that:
$$x^2 + y^2 = u^4 + v^4 + 2u^2v^2 = (u^2 + v^2)^2, \therefore \sqrt{x^2 + y^2} = u^2 +v^2$$
so we get the desired equation.