Suppose $X(u,v)=(u-2v,u+3v)=(x,y)$ and $F(x,y)=\left( f(x,y),g(x,y) \right)$ with $f(x,y),\: g(x,y)$ continuously differentiable. Express the derivative of $F(X)(u,v)$ in terms of derivatives of $x$ and $y$.
I am confused about the exercise. I don't know if I need to do the following:
\begin{align*} \frac{\partial }{\partial u}\left(u-2v\right)= 1\\ \frac{\partial }{\partial v}\left(u-2v\right)=-2 \\ \frac{d}{du}\left(u+3v\right)=1 \\ \frac{d}{dv}\left(u+3v\right)=3 \end{align*}
How did this come about:
\begin{align*} F(X)(u, v)=\left( f(u-2v,\: u+3v),\: g(u-2v,\: u+3v) \right) \end{align*}
Hint: Define $H(u,v)=(F\circ X)(u,v)=\big(f(X(u,v)), g(X(u,v))\big)^\intercal=(p(u,v),q(u,v))^\intercal$
By the chain rule $$\begin{align} H'(u,v)&= F'(X(u,v))X'(u,v)\\ &=\begin{pmatrix} \partial_xf(X(u,v)) & \partial_yf(X(u,v))\\ \partial_x g(X(u,v)) & \partial_yg(X(u,v)) \end{pmatrix}\begin{pmatrix} 1 & -2\\ 1 & 3 \end{pmatrix}\\ &=\begin{pmatrix} \partial_xf(X(u,v))+\partial_yf(X(u,v) & -2\partial_xf(X(u,v))+3\partial_yf(X(u,v))\\ \partial_xg(X(u,v))+\partial_yg(X(u,v) & -2\partial_xg(X(u,v))+3\partial_yg(X(u,v)) \end{pmatrix} \end{align} $$
From here, you can read out what $\partial_xp$, $\partial_yp$, $\partial_xq$ and $\partial_yq$ are, i.e. what $H'(u,v)$ is.