How to find $\frac{\partial f}{\partial x}(a,b)$ and $\frac{\partial f}{\partial y}(a,b)$ of a implicit function? I need help with this problem:

For each of the following functions $F:\mathbb{R}^3\rightarrow\mathbb{R}$, show that the equation $F(x,y,z)=0$ defines implicitly a countinuously differentiable function $z=f(x,y)$ in a neighbourhood of the given point $(a,b,c)$. Find  $\frac{\partial f}{\partial x}(a,b)$ and $\frac{\partial f}{\partial y}(a,b)$.
(a) $F(x,y,z)=x^3+y^3+z^3-xyz-2\quad\quad$    at $(1,1,-1)$
(b) $F(z,y,z)=x^2+z^3-z-xy\sin z\quad\quad\ \ \ $  at $(1,1,0)$
Answers: (a) $-2,-2$; (b) $1, 0$.

Ok, so for (a), I was able to show that $F$ defines a continuously differentiable function, but I don't know how to find the partial derivatives, this is what I did:


*

*$$F(1,1,-1)=1+1-1+1-2=0$$ $$\frac{\partial F}{\partial z}(x,y,z)=3z^2-xy\Rightarrow \frac{\partial F}{\partial z}(1,1,-1)=2\neq0$$ Thus, tthere exists a neighbourhood $N$ of $(1,1)$, a neighbourhood $M$ of $-1$ and a conitnuously differentiable function $f:N\subset\mathbb{R}^2\rightarrow\mathbb{R}$ such that:


*

*$f(1,1)=-1$ and $f(N)\subset M$.

*for each $(x,y)\in N$, $F(x,y,z)=0$ is solved uniquely by $z=f(x,y)$
Now, I don't know how to find the function $f$, since I cant $F(x,y,z)$ for $z$, how do I find $f$ in order to find the partial derivatives at $(a,b)$?


*I think that this part is wrong, since $F(1,1,0)=1+0-0-\sin(0)=1$, not $0$, am I right?

 A: You need to use a different set of relations for implicit functions, especially those that are not separable.
For an implicit function where $F(x, y, z) = 0$, we have:
$$\frac{\partial f}{\partial x} = -\frac{F_x}{F_z}$$
$$\frac{\partial f}{\partial y} = -\frac{F_y}{F_z}$$
So we have for (2):
$$\frac{\partial f}{\partial x}_{(1, 1, 0)} = -\frac{2x-y\sin z}{3z^2-1-xy\cos z}\bigg|_{(1, 1, 0)} = -\frac{2}{-1-1} = 1$$
$$\frac{\partial f}{\partial x}_{(1, 1, 0)} =-\frac{-x\sin z}{3z^2-1-xy\cos z}\bigg|_{(1, 1, 0)} = -\frac{0}{-1-1} = 0$$
So (b) is $(1, 0)$ as expected.

If you don't want to use the relations above, consider
$$F(x,y,z)=x^2+z^3-z-xy\sin z$$
Since $F(x, y, z) = 0$, $dF = 0$
$$\implies dF = \frac{\partial F}{\partial x}dx + \frac{\partial F}{\partial y}dy + \frac{\partial F}{\partial z}dz = 0$$
$$(2x - y\sin z - (3z^2 - 1 - xy\cos z)\frac{\partial z}{\partial x}) dx + (-x\sin z +(3z^2 - 1 - xy\cos z) \frac{\partial z}{\partial y})dy + \frac{\partial F}{\partial z}dz = 0$$
Since the above relation is exactly $0$, it follows that the coefficient of $dx$ is exactly $0$ and the coefficient of $dy$ is also exactly $0$. Hence:
$$2x - y\sin z - (3z^2 - 1 - xy\cos z)\frac{\partial z}{\partial x} = 0$$
$$-x\sin z +(3z^2 - 1 - xy\cos z) \frac{\partial z}{\partial y} = 0$$
These give rise to the same equations as earlier.
