Find the equation of the plane tangent to the surface at a given point I am given the equation of a surface:
$$x^3+y^3+z^3-3xyz =0$$
And I need to find the equation of the plane tangent to this surface at $(1,1,1).$ 
At first, this task did not look easy for me as we are not given an explicit equation of a surface, but I tried using implicit differentiation assuming that $z$ depends on $x$ and $y$, therefore: 


*

*Differentiate with respect to $x$
$$3x^2+3z^2 \frac {\partial z}{\partial x} - 3yz - 3xy \frac{\partial z}{\partial x} = 0$$ and I use this equation to solve for $\frac{\partial z}{\partial x}$

*Differentiate with respect to $y$:
$$3y^2 + 3z^2 \frac{\partial z}{\partial y} - 3xz - 3xy \frac{\partial z}{\partial y} = 0 $$
And I solve for $\frac{\partial z}{\partial y}$
Now, I have both partial derivatives which I can evaluate at $(1, 1, 1)$ to obtain the equation of the plane. 
Is it the correct way to solve this problem? 

The so called surface $f(x,y,z) = x^3+y^3+z^3-3 x y z= 0$ is the product of a plane and a null radius cylinder or
$$
f(x,y,z) = (x+y+z)(x^2+y^2+z^2-x y- y z - x z) = 0
$$
The plane doesn't contain the point $(1,1,1)$ and the cylinder contains it but it has null radius so it is reduced to a line (generatrix) and a line has not unique normal vector.
