find the total differential of this equation $ xyz + \sqrt{ x^2 + y^2 + z^2} = \sqrt 2 $ How to calculate the total differential of $ z= z(x,y)$, which is  $ xyz + \sqrt{ x^2  + y^2 + z^2} = \sqrt 2 $ at point (1, 0, -1)？
The evaluation of mine seems wrong,
$ dz= \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y}  dy = (yz+ \frac{x}{\sqrt{x^2  + y^2 + z^2}})dx + (xz+ \frac{y}{\sqrt{x^2  + y^2 + z^2}})dy  = \frac{dx}{\sqrt2} - dy $
Appreciate any helps.
 A: I have done the following workings for the question.  I believe the answer is ok but please forgive if it is incorrect (and point out my oversight if possible).  
As mentioned in the comments $z$ is implicitly a function of both $x$ and $y$.  To be specific $z=z(x,y)$
First let's calculate $\frac{\partial z}{\partial x}$ remembering the product and chain rules:
You have:
$$xyz+\left( x^2+y^2+z^2 \right)^{1/2}=\sqrt{2}\\
yz+xy\frac{\partial z}{\partial x}+\frac{1}{2}\left(2x+2z\frac{\partial z}{\partial x}\right)\left(x^2+y^2+z^2\right)^{-1/2}=0$$
Now substitute $(1,0,-1)$ which, after simplifying, leads to $$2-2\frac{\partial z}{\partial x}=0\implies\frac{\partial z}{\partial x}=1$$
The case for$ \frac{\partial z}{\partial y}$ plays out similarly: $$xz+xy\frac{\partial z}{\partial y}+\frac{1}{2} \left( 2y + 2z \frac{\partial z}{\partial y}\right) \left(x^2+y^2+z^2 \right)^{-1/2}=0$$
Now substitute $(1,0,-1)$ leading to $\frac{\partial z}{\partial y}=\sqrt{2}$
Now $dz=\frac {\partial z}{\partial x}dx+\frac{\partial z}{\partial y} dy \implies dz=dx+\sqrt{2}dy$
