Find the partial derivative of $z$ with respect to partial derivative of $x$ at point $(1,1,1)$ in equation $xy-z^3x-2yz = 0.$
If I am not mistaken, after simplification of the partial derivative, one may obtain $(y-3z^2-2z)\frac{dz}{dx}=0,$ after which $dz/dz = 0$? But I am not sure if this is the correct thought process.
I am just not sure if I took the partial derivative correctly.
You are mistaken. $$\frac{\partial}{\partial x}(xy -z^3x -2yz) = \frac{\partial}{\partial x}(0)$$ $$y -3z^2\frac{\partial z}{\partial x}x - z^3 -2y\frac{\partial z}{\partial x} =0$$ $$y - z^3 -(3xz^2 + 2y)\frac{\partial z}{\partial x} =0$$
$$\frac{\partial z}{\partial x} =\frac{y - z^3}{3xz^2 + 2y}$$
You are to consider all parameters apart from $x$ and $z$ fixed, that is, not variable. Then taking derivatives with respect to $x$ gives $$(xy-xz^3-2yz)'=0',$$ or $$(xy)'-(xz^3)'-2(yz)'=0,$$ which gives $$y(x)'-[x(z^3)'+z^3(x)']-2y(z')=0,$$ or $$y(1)-[x(3z^2z'+z^3(1)]-2yz'=0,$$ or $$y-3xz^2z'-z^3-2yz'=0,$$ from which you should have spotted where you went wrong.