## Proof that X is a smooth surface

I have the following problem:

At what point is
X = {(x,y,z) : (x^2 + y^2 + z^2)^2 - 4(x^2 + y^2) = 0}
a smooth surface? Give the equation of its tangent space at such a point.

Here's my attempt at a proof:
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I believe that this will be a 2-dimensional smooth surface if, for F(x,y,z) = (x^2 + y^2 + z^2)^2 - 4(x^2 + y^2), DF(x,y,z) has rank 1.

By my calculations,

dF/dx = 4x(x^2 + y^2 + z^2) - 8x
dF/dy = 4y(x^2 + y^2 + z^2) - 8y
dF/dz = 4z(x^2 + y^2 + z^2)

So DF(x,y,z) is a 1x3 matrix with those as entries. Now DF(x,y,z) will have rank 1 provided that it is not true that

dF/dx = dF/dy = dF/dz = 0

I believe that we can only have all partials equal zero at the points where (x^2 + y^2) = 2 and z = 0, or at the point (0,0,0).

This means that it is a smooth surface provided that those conditions are not met.
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Is this correct? If not, what did I mess up on? I can't see any obvious logical errors, but I'm not 100% sure of it.

Assuming this is correct, could anybody point me in the right direction for figuring out the tangent space?