Let $f: U \subset \mathbb{R}^2 \to \mathbb{R}$ be a continuous function in the open subset $U$ of $\mathbb{R}^2$ such that $$(x^2+y^4)f(x,y)+f(x,y)^3 = 1, \forall (x,y) \in U$$ Show that $f$ is of class $C^1$ in $U$.
I think that is an application of implicit function theorem, but I don't know how to solve it, because I only saw examples about system of linear equations.
On
As usual, Ted gave a great answer (and you should accept it). I'll just nitpick a bit, which you might or might not appreciate now: being $C^1$ is a local property, so you want to check that given $(x_0,y_0)\in U$, then $f$ is of class $C^1$ in a neighbourhood of $(x_0,y_0)$.
Great, so once you apply the IFT to the funcion $F$ in Ted's answer, you get a neighborhood $V$ of $(x_0,y_0)$ (which you can assume it's contained in $U$, replacing $V$ by $V\cap U$ if necessary) and an open interval $I$ centered in $z_0$ such that for all $(x,y) \in V$ there is a unique $\varphi(x,y)\in I$ such that $$F(x,y,\varphi(x,y))=1, $$and this implicit function $\varphi:V\to I$ is also $C^1$. By the uniqueness above, $f|_V= \varphi$ is $C^1$. And that's the reason $f$ is $C^1$. I just wanted to ilustrate the importance of the uniqueness part of the theorem, so you can use its full power without missing anything.
The implicit function theorem would apply directly to tell you that if you have the equation $$F(x,y,z) = (x^2+y^4)z + z^3 = 1,$$ then you can locally solve for $z=f(x,y)$ near $(x_0,y_0,z_0)$ as a $C^1$ function provided
(1) $F$ is $C^1$ (it is)
(2) $\partial F/\partial z \ne 0$ at the point $(x_0,y_0,z_0)$.
What do you notice about $\partial F/\partial z$ at every point in $U$?