Multivariable calculus - find derivative using implicit differentiation

Short simple question which i managed to solve partially.

we are given the equation $x^2+y^2-z^2+xz-yz-1=0$.

Show using the implicit function theorem that this equation sets in the neighborhood of $(1,0,1)$ $z$ as a function of $x$ and $y$ and find: $\frac{dz}{dx},\frac{dz}{dy},\frac{d^2z}{dxdy}$ where $x=1$ and $y=0$

My answer:

Define $F(x,y,z)=x^2+y^2-z^2+xz-yz-1$. Notice that $F(1,0,1)=0$.

$F_x=2x+z$ , $F_y=2y-z$ , $F_z=-2z+x-y$ are all continouos functions, and $F$ itself is continuous as well, and as such $F \in C^1$.

Also notice that $F_z(1,0,1)=-1 \neq 0$ and so $F_z$ is invertible at that point, and according to IFT, we can represent $z=z(x,y)$ as a function of $x$ and $y$ around that point.

Now where I am having difficulties.

if you look at an earlier question I asked Question about Implicit function theorem I tried deriving like I did then, and it does not work. If i was to do the same thing:

$$\frac{dz}{dx} = -\frac{F_x}{F_z} = -\frac{2x+z}{x-y-2z}$$

I didn't get rid of the $z$. and the question only said $x=1$ and $y=0$. Is that because I must put $z=1$? or should the derivative be "free" of $z$?

What you did is correct. You just need to set $(x,y,z)=(1,0,1)$ in the last expression. I suppose the difficulties you had arise from the informal way in which you solved things (for instance, not indicating at which point you're taking the partial derivatives).

I'll write up an answer similar to what I did in the linked question.

The IFT gives you the existence of neighborhoods $V$ of $(1,0)$ and $W$ of $1$ such that there exists a smooth function $G\colon V\to W$ with the property that $G(1,0)=1$ and $\forall (x,y)\in V\left(F(x,y,G(x,y))=0\right)$. (Here you - not just you, almost everyone - abuses the notation again by saying $z\equiv G$, but $z$ isn't a function).

Now let $H\colon V\to W, (x,y)\mapsto F(x,y,G(x,y))$ and take its partial derivatives $H_x$ and $H_y$:

• $0=H_x(x,y)=F_x\left(x,y, G(x,y)\right)+F_z\left(x,y,G(x,y)\right)G_x(x,y),$
• $0=H_y(x,y)=F_y\left(x,y, G(x,y)\right)+F_z\left(x,y,G(x,y)\right)G_y(x,y),$

for all $(x,y)\in V$, in particular it holds for $(1,0)$ yielding

• $G_x(1,0)=-\dfrac{F_x\left(1,0, G(1,0)\right)}{F_z\left(1,0,G(1,0)\right)}=-\dfrac 3{-1}=3$ and
• $G_y(1,0)=\dfrac{F_y\left(1,0, G(1,0)\right)}{F_z\left(1,0,G(1,0)\right)}=-\dfrac {-1}{-1}=-1$.

It remains to compute $\dfrac {\partial ^2G}{\partial x\partial y}(1,0)$ which I think, in subscript notation is denoted by $G_{yx}(1,0)$. To do this just differentiate $H_y$ with respect to $x$: for all $(x,y)\in V$ one has $$ 0=H_{yx}(x,y)=\left[F_{yx}\left(x,y, G(x,y)\right)+F_{yz}\left(x,y, G(x,y)\right)G_x(x,y)\right]+\left[\color{green}(F_{zx}\left(x,y,G(x,y)\right)+F_{zz}\left(x,y,G(x,y)\right)G_x(x,y)\color{green})G_y(x,y)+F_{z}\left(x,y,G(x,y)\right)G_{yx}(x,y)\right]. $$

Evaluate at $(1,0)$ and you get whatever you get.