Hallamos la matriz de derivadas de f.
Las funciones son
h: R3 ----> R
k: R3 ----> R
(h,k): R3 ----> R2
F: R2 ----> R
f = Fo(h,k): R3 ----> R
Aplicamos la regla de la cadena.
$$\begin{pmatrix}
\frac{\partial f}{\partial x}&\frac{\partial f}{\partial y}&\frac{\partial f}{\partial z}
\end{pmatrix}
=
\begin{pmatrix}
\frac{\partial F}{\partial h}&\frac{\partial F}{\partial k}
\end{pmatrix}
\begin{pmatrix}
\frac{\partial h}{\partial x}&\frac{\partial h}{\partial y}&\frac{\partial h}{\partial z}\\
\frac{\partial k}{\partial x}&\frac{\partial k}{\partial y}&\frac{\partial k}{\partial z}
\end{pmatrix}=
\\
.
\\
\begin{pmatrix}
\frac{\partial F}{\partial h}\frac{\partial h}{\partial x}+\frac{\partial F}{\partial k}\frac{\partial k}{\partial x}\quad &
\frac{\partial F}{\partial h}\frac{\partial h}{\partial y}+\frac{\partial F}{\partial k}\frac{\partial k}{\partial y}\quad&\frac{\partial F}{\partial h}\frac{\partial h}{\partial z}+\frac{\partial F}{\partial k}\frac{\partial k}{\partial z}
\end{pmatrix}
\\
·
\\
\nabla f=\begin{pmatrix}
\frac{\partial F}{\partial h}\frac{\partial h}{\partial x}+\frac{\partial F}{\partial k}\frac{\partial k}{\partial x}, &
\frac{\partial F}{\partial h}\frac{\partial h}{\partial y}+\frac{\partial F}{\partial k}\frac{\partial k}{\partial y},&\frac{\partial F}{\partial h}\frac{\partial h}{\partial z}+\frac{\partial F}{\partial k}\frac{\partial k}{\partial z}
\end{pmatrix}$$
Y eso es todo.