Let $F(x,y,z)= f(x^2-y^2+z,2x-y+z)$ with $f: \mathbb R^2\rightarrow \mathbb R$ differentiable
a)Suppose $\frac{\partial f}{\partial v}(0,3)=-2$, $\frac{\partial F}{\partial w}(0,3)=1$ with $v=(1,1)$ and $w=(1,-2).$ Find $\nabla (0,3).$
b)Using the chain rule find $\nabla F(1,2,3)$
c) Find the tangent plane to Gr(F) in $(c, F(c))$ with c=(1,2,3)
Using the equivalence between directional derivates and the differential I got $\nabla f(0,3)=(-1,-1)$
My problem is how to solve b), I'm confused and don't know how to use the chain rule. My aproach:
Let $h(x,y,z)=(x^2-y^2+z,2x-y+z)$ then $$D F(1,2,3) = D(f\circ h)(1,2,3)$$ $$= Df(h(1,2,3)) \circ Dh(1,2,3)$$ $$=Df(0,3) \circ D h(1,2,3)$$
And $$Dh(1,2,3)= \begin{bmatrix} 2x& -2y & 1\\ 2& -1 & 1 \end{bmatrix} = \begin{bmatrix} 2& -4 & 1\\ 2& -1 & 1 \end{bmatrix}$$
thus $\nabla F(1,2,3)= (-4,5,-2)$ Is it right?