When is $d^{2} x$ =0? We want to find $d^{2} z$ :
$$
(4 x-3 z-16) d x+(8 y-24) d y+(6 z-3 x+27) d z =0
$$
So, in my book, we apply the differential operator d to the above equation :
We use the product rule ;
$$
(4 d x-3 d z) d x+(8 d y) d y+(6 d z-3 d x) d z+(6 z-3 x+27) d^{2} z=0
$$
So my question is :
Where did $d^{2} x$ and $d^{2} y$ go?
Are they equal to 0?
If so why
$d^{2} z$ is not 0?
Edit:
That's the problem :
So above $d^{2} z$ isn't asked in the question but it's needed to find the asked ones ;
Problem 1 (65 points)
We consider the function
$$
F(x, y, z)=2 x^{2}+4 y^{2}+3 z^{2}-3 x z-16 x-24 y+27 z+94
$$
and let $z=z(x, y)$ be an implicit function defined by the equality $F(x, y, z)=-1$.
1.1. Calculate $\frac{\partial z}{\partial x}, \frac{\partial z}{\partial y}, \frac{\partial^{2} z}{\partial x^{2}}, \frac{\partial^{2} z}{\partial y^{2}}$ and $\frac{\partial^{2} z}{\partial x \partial y}$ at point $(x=1, y=3, z=-5)$.
 A: With the context, you have $d^2 x=d^2y=0$, but $d^2 z$ has to be seen as $d^2 z(x,y)$, so that it's not $0$. Here,
$$dz=\frac{\partial z}{\partial x}dx+\frac{\partial z}{\partial y}dy$$
and thus
$$d^2 z =\left(\frac{\partial^2 z}{\partial x^2}dx
+\frac{\partial^2 z}{\partial y\partial x}dy\right)dx+\left(\frac{\partial^2 z}{\partial x\partial y}dx
+\frac{\partial^2 z}{\partial y^2}dy\right)dy.$$
A: You are forgetting to use the product rule.  If I have $d(8q\,dq)$, then the resulting differential will be $8\,dq^2 + 8q\,d^2q$.
$$ d\left((4 x-3 z-16)\, dx +(8 y-24)\, dy +(6 z-3 x+27) \,dz\right) =d( 0) $$
This yields
$$ (4x\,dx - 3z\,dz)\,dx + (4 x-3 z-16)\,d^2x + (8\,dy)\,dy + (8y-24)\,d^2y + (6\,dz - 3\,dx)\,dz + (6 z-3 x+27 )d^2z$$
Now, this only makes sense if you use the algebraically-manipulable version of the second derivate.  In this, the second derivative of $y$ with respect to $x$ is:
$$\frac{d^2y}{dx^2} - \frac{dy}{dx}\frac{d^2x}{dx^2}$$
This is described more fully in my paper, "Extending the Algebraic Manipulability of Differentials".
A: I think that part of the misunderstanding of some people is that you need to say what the notation of $dz$ means. I don't recommend solving these problems using that if you are not very familiar with it (otherwise, see SacAndSac's answer).
In any case, to solve the problem you can always take the partial derivatives of the equality $$F\big(x,y,z(x,y)\big)=2x^2+4y^2+3z^2−3xz−16x−24y+27z+94=-1$$ and compute what you want.
To compute $\partial z / \partial x$ take the partial derivative $\partial/\partial z$ in the equality $F(x,y,z)=-1$, but remember that $z$ is a function that depends on $x$ and $y$: $z=z(x,y)$.
$$ \frac{\partial F}{\partial x}(x,y,z)=4x+0+6z\frac{\partial z}{\partial x}−3z-3x\frac{\partial z}{\partial x}−16+0+27\frac{\partial z}{\partial x}+0=0.$$
Now you simply use the extra data of the point where you are asked to compute the derivatives, in this case $(x,y,z)=(1,3,-5)$.
$$4·1+6·(-5)\frac{\partial z}{\partial x}−3·(-5)-3·1\frac{\partial z}{\partial x}−16+27\frac{\partial z}{\partial x}=0.$$
Isolate the derivative and you are done:
$$ \frac{\partial z}{\partial x}=-\frac{4+15-16}{-30-3+27}=\frac{1}{2}.$$
You can do the same for the remaining partial derivatives of $z$.
