$F$ is differentiable at $x_0$, but $S=\{ x \in \mathbb{R}^3 : F(x)=c\}$ is not smooth at $x_0$ 
Consider the surface defined implicitly by $x^3 - x^2y^2 + z^2 = 0$. Does it have a well-defined tangent plane at the origin? Why or why not?

The polynomial function 
$$F(x,y,z)=x^3 - x^2y^2 + z^2$$ 
is $C^1$ in its domain; in particular, it's differentiable at the origin. I thought that this was sufficient to claim that its level set at height $0$ is smooth and has a well-defined tangent plane at the origin. But it's not. Is this simply because $\nabla F(0,0,0)=0$? 
If that wasn't the case, the Implicit Function Theorem would imply that $S$ is the graph of some $C^1$ function near the origin, right?
So, to be extremely methodical when answering theses questions, 
(1) If $\nabla F (x_0) \neq 0$, then the level set is smooth at $x_0$ iff $F$ is $C^1$ at $x_0$;
(2) If $\nabla F (x_0) = 0$, we have to do algebra on the equation that defines the level set...? Or what?
 A: In fact the problem is whether a level-set $S(c) = F^{-1}(c)$ is locally a $2$-dimensional $C^1$-submanifold of $\mathbb R^3$ at $x_0 \in S(c)$. This is guaranteed if $\nabla F(x_0) \ne 0$.
If $\nabla F(x_0) = 0$, we can say that $c$ is not a regular value of $F$ and frequently $S(c)$ will not be locally a $2$-dimensional $C^1$-submanifold at $x_0$ and thus not have a well-defined tangent plane at $x_0$. However, this depends on $F$. As an example look at $F : \mathbb R^3 \to \mathbb R, F(x,y,z) = z^2$. The level-set $S(0)$ is the $x$-$y$-plane which clearly is its own tangent plane at each point $x_0 \in S(0)$. But certainly $\nabla F(0) = 0$.
Thus we have to look at the set in your question. 
Its intersection $S_h(0)$ with the plane $z = h$ (which is parallel to the $x$-$y$-plane) consists of all $(x,y,h)$ with $x^2y^2 = x^3 + h^2$. We shall omit $h$ and only consider $(x,y)$ to understand what this curve looks like.
For $h=0$ the set $S_0(0)$ is the union of the parabola $x = y ^2$ and the line $x = 0$. For $h \ne 0$ it is impossible that $x =0$ and we get $y^2 = x + (h/x)^2$. The curve $S_h(0)$ has three connected components which look as follows (for $h=1$):

As $x \to 0$, we have $y \to \pm\infty$.
As $h \to 0$, the component left of the $y$-axis approaches the $y$-axis, similarly the components right of the $y$-axis approach the upper and lower half of the set $S_0(0)$.
Thus we can imagine what $S(0)$ looks like and we see that it is not a surface in the usual sense (just look at $S_0(0)$). It is not a $2$-dimensional $C^1$-submanifold of $\mathbb R^3$ at $0$. It therefore does not have a well-defined tangent plane at the origin.
