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.
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.