If $x=u^3+v^3$ and $y=u*v-v^2$ are solved for $u$ and $v$ in terms of $x$ and $y$, evaluate $\partial u/\partial x$ at the point where $u = 1$ and $v = 1$.
It also asks me to evaluate $\partial u/\partial y$, $\partial v/\partial x$, $\partial v/\partial y$, and $\partial (u,v)/\partial (x,y)$ for the same $u = 1$ and $v = 1$.
I am not sure at all what to do here. I did try and get $u$ and $v$ on one side, but the expression is so ridiculously long that I don't think I did it right... help is very much appreciated!
At the point $p:=(1,1)$ the map $$f:\quad (u,v)\mapsto\left\{\eqalign{x&=u^3+v^3\cr y&=uv-v^2\cr}\right.$$ is regular, because the determinant of the Jacobian $$J_f(u,v)=\left[{\partial(x,y)\over\partial(u,v)}\right]=\left[\matrix{3u^2&3v^2\cr v&u-2v\cr}\right]$$ at $p$ is given by $$\det J_f(p)=\det\left[\matrix{3&3\cr 1&-1\cr}\right]=-6\ne0\ .$$ According to the implicit function theorem this implies that $f$ maps a neighborhood $U$ of $p$ bijectively onto a neighborhood $V$ of the point $q:=f(p)=(2,0)$, and the inverse map $g:\>V\to U$ is again differentiable at $q$. Furthermore $$dg(q)=\bigl(df(p)\bigr)^{-1}\ ,$$ hence the Jacobian of $g$ at $q$ is given by $$J_g(q)=\bigl(J_f(p)\bigr)^{-1}=\left[\matrix{3&3\cr 1&-1\cr}\right]^{-1}=\left[\matrix{{1\over6}&{1\over2}\cr {1\over6}&-{1\over2}\cr}\right]\ .$$ The last matrix is already the matrix $$\left[\matrix{u_x&u_y\cr v_x&v_y\cr}\right]_q\ ,$$ and this gives the answers to all your questions.