Let $X$ be the vector field on $\mathbb{R^3}$-$(0,0,0)$ defined by $$X=y^2\frac{\partial}{\partial{x}}+z\frac{\partial}{\partial{y}}-x\frac{\partial}{\partial{z}}$$
and let $F:\mathbb{R^3}$-$(0,0,0)→(0,0,0)$ be the mapping defined by $$F=(u,v,w)$$where $u=xy$, $v=yz$ and $w=xz$.
If $$F_*(X)=A\frac{\partial}{\partial{u}}+B\frac{\partial}{\partial{v}}+C\frac{\partial}{\partial{w}}$$
then find $A,B,C$. Here $F_*$ is the differential of $F$, that it $F_*=dF$.
My approach to this question was writing $\partial{u}=\partial{(xy)}$ and so on. Is there a such thing that $\partial{(xy)}$? When I look up to definitions of partial derivatives I couldn't find anything like to this notation. I guess I cannot apply $d(xy)=xdy+ydx$ to the partials.
I'm so lost...Any advice/solution will be appreciated.
On one side,
$$dF = {\partial F \over \partial x} dx + {\partial F \over \partial y} dy + {\partial F \over \partial z} dz.$$
On the other,
$$dF = {\partial F \over \partial u} du + {\partial F \over \partial v} dv + {\partial F \over \partial w} dw$$,
and since $$du = {\partial u \over \partial x} dx + {\partial u \over \partial y} dy + {\partial u \over \partial z} dz$$ (and analogous relations exist for $v$ and $w$), subsitute these in the latter expression for dF and then factor out $dx$, $dy$, and $dz$ to get $A$, $B$, and $C$.