OK, so I'm out of ideas. The question is: Let f(x,y) be a function with continous partial derivatives that upholds:
$$ \bigtriangledown f (0,-8) = (-3\widehat{i}, 5\widehat{j}) $$ let g(x,y) be the function : $$ g(x,y) = f(xy+x^2 , xy-y^2) $$
Calculate: (a Vector) $$ \bigtriangledown g(-2,2) = ? $$
My take on this: $$ \bigtriangledown g = \left(\frac{\partial \:g}{\partial \:x},\:\frac{\partial \:g}{\partial \:y}\right) $$ let u,v be: $$ u\left(x,y\right)\:=\:xy+x^2\:\:,\:v\left(x,y\right)\:=\:xy-y^{2\:} $$
and then By the chain Rule: $$ \frac{\partial g}{\partial x}\:=\:\frac{\partial g}{\partial u}\:\cdot \:\frac{\partial u}{\partial x}\:+\:\frac{\partial g}{\partial v}\cdot \frac{\partial v}{\partial x} $$ $$ \frac{\partial g}{\partial y}\:=\:\frac{\partial g}{\partial u}\:\cdot \:\frac{\partial u}{\partial y}\:+\:\frac{\partial g}{\partial v}\cdot \frac{\partial v}{\partial y} $$
Placing du/dx du/dy and dv/dx dv/dy , and evaluating at the point (-2,2) $$ \frac{\partial g}{\partial y}\:=\:+2\frac{\partial g}{\partial u}\:+6\:\frac{\partial g}{\partial v} $$ $$ \frac{\partial g}{\partial x}\:=\:+2\frac{\partial g}{\partial u}\:-2\:\frac{\partial g}{\partial v} $$ My question is how to procceed further?, What do i do with The partial derivatives of g with respect to u and v.
I would think that you could use the chain rule on its matrix form. Let
$$h(x,y)=(xy+x^2, xy-y^2)$$
and observe that $h(-2, 2)=(0,-8)$. Now $g(x,y)=(f \circ h)(x,y)$ and according to the chain rule
$$(\nabla (f \circ h)(x,y))^T = (\nabla f(h(x,y))^T J_h (x,y).$$
So we get
\begin{align} (\nabla (f \circ h)(-2,2))^T &= (\nabla f(0,-8))^T J_h (-2,2) \\ &= [-3 \quad 5] \begin{bmatrix} -2 & -2 \\ 2 & -6 \end{bmatrix} = [16 \quad -24] \end{align}
Now, I don't think this is the most elegant way of using the chain rule here, so I would like to see other solutions.