# How to prove $f$ is $C^\infty$

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.

• Not sure if this helps, but that implies $f$ satisfies the functional equations $$\begin{split} \frac{1}{f(x,y)} &= x^2+y^4 + f(x,y)^2\\ f(x,y) &= \frac{1}{x^2+y^4 + f(x,y)^2} \end{split}$$ – gt6989b Apr 17 at 15:01
• Yeah, we can see that $f(x,y)\ne 0$ otherwise $f$ would not satisfy the equation. – ipreferpi Apr 17 at 15:04
• Idea: Implicit Function Theorem. – Martín-Blas Pérez Pinilla Apr 17 at 15:11
• Idea: Cardano's formula. – Michael Hoppe Apr 17 at 15:14
• More legible: $$f(x,y)= \frac{\left(\sqrt3 \sqrt{4 a^3 + 27} + 9\right)^{1/3}}{2^{1/3} 3^{2/3}}-\frac{(2/3)^{1/3} a}{\left(\sqrt3 \sqrt{4 a^3 + 27} +9\right)^{1/3}}$$ where $a=x^2+y^4$. – Michael Hoppe Apr 17 at 15:30

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.