Showing that $f$ is $C^\infty$ Question:

Let $f: U \to \mathbb R$ be a continuous function, with $U \subset \mathbb R^2$ open, 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 $C^\infty$.

Attempt:
Define $F(x,y,z) = (x^2 + y^4)z + z^3 - 1$. Then $$F_z(x,y,z) = (x^2 + y^4) + 3z^2 > 0$$
for any $(x,y,z) \in \mathbb R^3- \{0\}$, and if we fix $(x_0,y_0) \in U$, such that $(x_0,y_0) \neq 0$, then taking $z_0 = f(x_0,y_0) \in \mathbb R$, thus it follows that $F(x_0,y_0,z_0) = 0$. 
Now by the Implicit Function Theorem we have that $z$ is a defined as a function of $x$ and $y$, such that $$F(x,y,z) = F(x,y,\xi (x,y)) = 0 \tag{1}$$ for every $(x,y,z) \in B \times J$, here $B \subset \mathbb R^2$ and $J \subset \mathbb R$. Clearly $F$ is of class $C^\infty$ then $z = \xi (x,y)$ is of class $C^\infty$. 
As $f$ is continuous there exists $\delta > 0$ sufficiently small such that $f(B) \subseteq J$, then we may conclude by $(1)$ and by hypothesis $F(x,y,f(x,y)) = 0$, that  $f(x,y) = \xi(x,y)$, for every $x \in B$, it follows then that $f$ is of class $C^\infty$. 
 A: For $t\ge 0$ the function $z \mapsto z^3 + t z -1$ is strictly increasing, with limit $+ \infty ( -\infty ) $ at $ + \infty ( - \infty)$. Therefore, for every fixed $t \ge 0$ the equation $z^3 + t z - 1 = 0$ has a unique solution $z(t)$. Note that $0 < z(t) \le 1$ for every $t \ge 0$ and $t \mapsto z(t)$ is strictly decreasing.  For instance, $z(0) = 1$. Let's show that $t \mapsto z(t)$ is a $C^{\infty}$ function. Indeed, whenever we have $z^3 + t z - 1 = 0$ the derivative $\frac{\partial (z^3 + t z - 1) }{\partial z} = 3 z^2 + t >0$ so it's not zero. Using the implicit function theorem we conclude that around any $t\ge0$ the unique solution $z(t)$ is a smooth function. Hence $[0, \infty)\ni t \mapsto z(t) \in  (0, 1]$, being locally smooth, is smooth.
It's worth noticing that we have an explicit formula ( the equation being of degree $3$)
$$z(t) = \left( \sqrt{ \frac{1}{4} + \frac{t^3}{27}} + \frac{1}{2} \right )^{\frac{1}{3}} -\left( \sqrt{ \frac{1}{4} + \frac{t^3}{27}} - \frac{1}{2} \right )^{\frac{1}{3}}$$
Now we get $f(x,y) = z[ x^2 + y^4]$, a composition of smooth functions, hence smooth.
Let's point out that unless we specify some continuity for $f(x,y)$, we need to use the uniqueness of the real solution of some equation. For instance, the functional equation $f(x,y)^2 - (x^2 + y^4+1)$ also has dincontinuous solutions, since $z^2 - t=0$ has two solutions $\pm \sqrt{t}$ for $t > 0$
A: The discriminant of $P(z) = t z + z^3 - 1$ is $-4 t^3 - 27$, which is nonzero for $t \ge 0$.  Thus $\dfrac{\partial}{\partial z} P(z) \ne 0$ whenever $P(z) = 0$ with $t \ge 0$.  In particular, this applies when $t = x^2 + y^4$.
