TITLE: How to prove $f$ is $C^\infty$
QUESTION [6 upvotes]: Suppose $f:U \subseteq \mathbb{R}^2 \rightarrow \mathbb{R}$ is  continous  and 
  $$(x^2+y^4)f(x,y)+(f(x,y))^3=1 \: \text{for all} \: (x,y) \in U. $$  Prove $f$ is $C^\infty$.

This kind of exercise is new to me and I don't really have any idea how to derive that the derivative exist 
infinitely and it's continous.

REPLY [3 votes]: The answers really are in the comments to the main question, and are due to Martin Blas Perez and Michael Hoppe. Hence this post is CW. 
The function $z=f(x, y)$ satisfies the equation 
$$
(x^2+y^4)z+z^3=1, $$
so it can never vanish. In particular, by the implicit function theorem, $f$ must be $C^1$ and 
$$
\begin{array}{cc}
\frac{\partial z}{\partial x} = -\frac{2xz}{x^2+y^4+3z^2}, & \frac{\partial z}{\partial y} =-\frac{4y^3z}{x^2+y^4+3z^2},
\end{array}
$$
and we remark that the denominators can never vanish. By iterating this process we see that $f$ is infinitely differentiable. 
Actually, Micheal even computed an explicit expression for $z$. See his comment to the main question, which I hope he turns into an answer. From that expression it is manifest that $z$ is smooth.