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 $(x=1, y=3, z=-5)$.

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".
