$(x^2+y^4)f(x,y)+f^3(x,y)=1 \implies f\in C^1(\Omega)$ Let $\Omega\subset\mathbb R^2$ be open and $f:\Omega\to\mathbb R$ be a continuous function such that $$(x^2+y^4)f(x,y)+f^3(x,y)=1$$ for each $(x,y)\in \Omega$. Prove that $f\in C^1(\Omega)$.
If is assumed that $f$ is differentiable, easily we get that the partials derivatives are continuous in $\Omega$, so $f\in C^1(\Omega)$. But I don't know how to prove that $f$ is differentiable. 
 A: For every $(x,y)\in \mathbb R^2,$ there is a unique $z\in \mathbb R$ such that
$$\tag 1 (x^2+y^4)z + z^3 = 1.$$
That's because for each fixed $a\ge 0,$ the function $t\to at+t^3$ is a bijection of $\mathbb R$ onto $\mathbb R.$ Thus $(1)$ defines $z$ as a function of $(x,y)$ for $(x,y)\in \mathbb R^2.$ Let's call this function $g(x,y).$ 
Set $h(x,y,z) = (x^2+y^4)z + z^3.$ Note that 
$$\tag 2\frac{\partial h}{\partial z}(x,y,z) = x^2+y^4+ 3z^2.$$
This is nonzero except at $(x,y,z)=(0,0,0).$ Since $(0,0,0)$ doesn't solve $(1),$ we have $(2)$ non-zero at every point of the surface given by $(1).$ Because $h$ is $C^1,$ the implicit function theorem shows $g(x,y)$ is locally $C^1,$ hence globally $C^1.$ Since $f(x,y) = g(x,y)$ for $(x,y)\in \Omega$ by uniqueness, we have $f\in C^1(\Omega)$ as deisred. (It appears that $f\in C^\infty(\Omega )$ actually.)
A: By Cardano's formula
one has
$$f \left(x , y\right) = g \left(x , y\right)-\frac{{x}^{2}+{y}^{4}}{3 g \left(x , y\right)}$$
with
$$g \left(x , y\right) = \sqrt[\large{3}]{\frac{1}{2}+\sqrt{\frac{1}{4}+\frac{{\left({x}^{2}+{y}^{4}\right)}^{3}}{27}}}$$
It follows that $f \in  {\mathscr{C}}^{\infty } \left({\Omega}\right)$
