Find the derivate $f',f''$ of the implicit function $z=f(x,y)$ defined by the following equation:
$$F(x,y,z)=x^2+y^2+z^2-a^2=0$$
So the first step to build the Jacobi-Matrix $f'$ lead me to this: $$f'(x,y,z)=\begin{pmatrix} -\frac{x}{z} & -\frac{y}{z} \end{pmatrix}$$
I'm not sure in ranking the Jacobi-Matrix for $f''$: First I derivated partial: $$\partial_{xx} =-\frac{1}{z}, \partial_{yy}=-\frac{1}{z}, \partial_{xy}=\partial_{yx}=-1$$
For $\partial_z$ there isn't a function to derivate in fact of $f'$. So I got this: $$f''(x,y)=\begin{pmatrix} -\frac{1}{z} & -1 \\ -1 & -\frac{1}{z} \end{pmatrix}$$
Any hints?
One way to do it is to just use the chain rule:
$$F(x,y,f(x,y))=x^2+y^2+[f(x,y)]^2-a^2=0$$
Differentiate with respect to $x$:
$$2x+2ff_x=0\qquad (1)$$
Differentiate with respect to $y$:
$$2y+2ff_y=0\qquad (2)$$
From these you get $f_x=-x/z$ and $f_y=-y/z$ as you say. (You could also use the formula from the implicit function theorem to get this result.)
To get $f_{xx}$, just differentiate $f_x=-x/(f(x,y))$ with respect to $x$:
$$f_{xx}=-\frac{f-xf_x}{f^2}$$
and then substitute $f=z$ and $f_x=-x/z$:
$$f_{xx}=-\frac{z+x^2/z}{z^2}=-\frac{x^2+z^2}{z^3}.$$
You can find the other derivatives in a similar way. It is pretty tedious.