Use Implicit Differentiation to compute the partial derivate $\frac{\partial z}{\partial x}$ at (1,1) The question I am working on is: The equation $xy+z^3x-2yz=0$ defines z as a function x,y around the point (1,1,1). Use Implicit Differentiation to compute the partial derivate $\frac{\partial z}{\partial x}$ at (1,1).
This is the work I have done:
$\frac{\partial}{\partial x}(xy+z^3-2yz = 0)$
$\implies y + 3z^3\frac{\partial z}{ \partial x}$ + $z^3x -2y\frac{\partial z}{\partial x} = 0 \implies \frac{\partial z}{\partial x} = \frac{-y-z^3x}{3z^2-2y}$ 
Then evaulated at (1,1) $\frac{\partial z}{ \partial x} = \frac{-1-z^3}{3z^2-2}$
I'm not sure if this is correct, if someone can verify it for me that would be great. If it is incorrect please tell me what went wrong so I can fix it. Thank You.
 A: $$\begin{align}
\frac{\partial}{\partial x}(xy+z^3-2yz) &=y+3z^2\frac{\partial z}{\partial x}-2y\frac{\partial z}{\partial x}\\\\
&=y+(3z^2-2y)\frac{\partial z}{\partial x}\\\\
&=0\\\\
&\implies\bbox[5px,border:2px solid #C0A000]{\frac{\partial z}{\partial x}=-\frac{y}{3z^2-2y}}
\end{align}$$
At the point $(1,1,1)$, we have
$$\bbox[5px,border:2px solid #C0A000]{\frac{\partial z}{\partial x}=-1}$$
A: Notice, we have $$xy+z^3-2yz=0$$ Now, partially differentiating both the sides of above equation w.r.t. $x$ assuming $y$ to be constant, we get $$\frac{\partial }{\partial x}(xy+z^3-2yz)=\frac{\partial }{\partial x}(0)$$ 
$$\frac{\partial }{\partial x}(xy)+\frac{\partial }{\partial x}(z^3)-\frac{\partial }{\partial x}(2yz)=0$$
$$y\frac{\partial }{\partial x}(x)+3z^2\frac{\partial }{\partial x}(z)-2y\frac{\partial }{\partial x}(z)=0$$ 
$$y+3z^2\frac{\partial z}{\partial x}-2y\frac{\partial z}{\partial x}=0$$
$$-\frac{\partial z}{\partial x}(2y-3z^2)=-y$$
$$\color{red}{\frac{\partial z}{\partial x}}=\color{blue}{\frac{y}{2y-3z^2}}$$
Hence, at the point $(1, 1, 1)$, substituting $x=1, y=1, z=1$ we get
$$\color{red}{\frac{\partial z}{\partial x}}|_{(1, 1, 1)}=\color{blue}{\frac{1}{2(1)-3(1)^2}}$$
$$\color{red}{\frac{\partial z}{\partial x}}|_{(1, 1, 1)}=\color{blue}{-1}$$
