Doesn't $x^3+2y^3+3z^3=0$ give a surface in $R^3$? In my last exam on Advanced Calculus (following Spivak's Calculus on Manifolds), I couldn't solve the following question.

True or false: the set $S$ in $R^3$ given by $x^3+2y^3+3z^3=0$ is a surface.

Thanks in advance for any ideas!

What I know: $S$ minus $0$ is a surface, because it is the inverse image of a regular value; so $0$ is the only possible problem. Now, there is a neighbourhood of any point in a surface which is the graph of a differentiable function in two variables. Moreover, the functions $-(1/3)(x^3+2y^3)^{1/3}$, $-(1/2)(x^3+3y^3)^{1/3}$, and $-(2y^3+3z^3)^{1/3}$ are not differentiable at $(0,0)$.
So, right  now my guess is: false.
 A: You are correct that $S$ is not a smooth surface in $\mathbb{R}^3$.  Here is a rough sketch of a proof.
Lemma 1:  $S$ is a surface iff $T = \{(x,y,z)\in \mathbb{R}^3: x^3 + y^3 + z^3 =0 \}$ is a surface.
Proof idea:  Consider $f:\mathbb{R}^3\rightarrow \mathbb{R}^3$ with $f(x,y,z) = (x, \sqrt[3]{2}y, \sqrt[3]{3}z)$.  Then $f$ is a diffeomorphism, and therefore maps surfaces to surfaces.  But $f(S) = T$ and $f^{-1}(T) = S$.
Note also that $f(0,0,0) = (0,0,0)$, so $(0,0,0)$ is the only potential bad point of $T$.
Lemma 2:  If $(x,y,z)\in T$ and $\lambda\in\mathbb{R}$, then $\lambda(x,y,z)\in T$.
Proof:  ommitted
Lemma 3:  Suppose a surface contains a straight line $\ell$.  Then for any $p\in \ell$, the tangent plane to the surface at $p$ contains $\ell$.
Proof:  One way of building the tangent plane is by picking two curves in your surface going through the point, then taking their derivatives at the point.  Just make sure to pick $\ell$ as one of your curves.
Lemma 4:  $T$ contains (at least) three lines $\ell_1, \ell_2, \ell_3$ through the origin which are not contained in any single plane.  Hence, $T$ cannot have a tangent plane at the origin, so $T$ is not a surface.
Proof:  The points $p_1 = (1,-1,0)$, $p_2 =(1,0,-1)$, and $p_3 = (1,1,-\sqrt[3]{2})$ are all in $T$.  By Lemma 2, so are the lines $\ell_i = p_i\cdot t$.  Their derivatives are $p_1, p_2,$ and $p_3$, thought of as vectors.  These three vectors area clearly independent.  More specifically, one can easily verify that $\ell_3$ is not contained in the (unique) plane made by $\ell_1$ and $\ell_2$.  For example, one can compute the cross product of $p_1$ and $p_2$ (that is, the normal vector of the plane containing $\ell_1$ and $\ell_2$) and verify that $p_3$ is not perpendicular to this cross product.
