Find all points on a surface which have a tangent plane parallel to given plane - is my method correct? The question given is to find all points on the surface given by $x^3 - y^3 - 2xy - z = 0$ which have a tangent plane which is parallel to $6x - 6y - z = 0$. 
So, I found the two gradient vectors (which we defined as $f_{x}e_{1} + f_{y}e_{2} + f_{z}e_{3}$) and said that one must be a scalar multiple of the other in order to be parallel. Then, as $f_{x} = 6k$ and $f_{y} = -6k$, I was able to set $f_{x} = -f_{y}$ and by cancelling I solved to get $y = x - \frac{2}{3}$ and worked out that all points of the form $(x, x - \frac{2}{3}, z)$ appear to satisfy the conditions given. Then I substituted this into the surface equation to get a value for $z$ which simplified nicely to $\frac{8}{27}$. So, all points of the form $(x,x - \frac{2}{3}, \frac{8}{27})$ is the final answer I got.  I have tried a couple of values for $x$ and the gradient vector does work out to be a scalar multiple of the plane's gradient. 
I would really appreciate if somebody could quickly work out the answer and verify if I have found the correct answer, or tell me if I am doing it wrong?  
Thanks! Helen
 A: $(3x^2-2y,-3y^2-2x,-1)=(6,-6,-1)$, since their normals must also be parallel as well. This will give $(x+y)(3(x-y)-2)=0$ solving this you can get the result.
A: Given the surface
$$z = f(x,y) = {x^3} - {y^3} - 2xy$$
we have a parametrization:
$$\phi (x,y) = (x,y,{x^3} - {y^3} - 2xy)$$
with partials:
$$\begin{gathered}
  {\partial _x}\phi  = \left( {1,0,3{x^2} - 2y} \right) \hfill \\
  {\partial _y}\phi  = \left( {0,1, - 3{y^2} - 2x} \right) \hfill \\ 
\end{gathered} $$
and an normal vector:
$${\partial _x}\phi  \times {\partial _y}\phi  = \left( {2y - 3{x^2},2x + 3{y^2},1} \right)$$
Normal vector from the plane and normal vector must be parallel,
so we may set:
$$\left( {2y - 3{x^2},2x + 3{y^2},1} \right) = \left( { - 6,6,1} \right)$$
This leads to the system:
$$\begin{gathered}
  2y - 3{x^2} =  - 6 \hfill \\
  2x + 3{y^2} = 6 \hfill \\ 
\end{gathered} $$
with four solutions:
$$\begin{gathered}
  (x,y) = \left( {\frac{1}{3} - \sqrt {\frac{5}{3}} , - \frac{1}{3} - \sqrt {\frac{5}{3}} } \right) \hfill \\
  (x,y) = \left( {\frac{1}{3} + \sqrt {\frac{5}{3}} , - \frac{1}{3} + \sqrt {\frac{5}{3}} } \right) \hfill \\
  (x,y) = \left( {\frac{1}{3}\left( { - 1 - \sqrt {19} } \right),\frac{1}{3}\left( {1 + \sqrt {19} } \right)} \right) \hfill \\
  (x,y) = \left( {\frac{1}{3}\left( { - 1 + \sqrt {19} } \right),\frac{1}{3}\left( {1 - \sqrt {19} } \right)} \right) \hfill \\ 
\end{gathered}$$
