Prove that there exists a function $f$ such that $\forall x \in \mathbb{R}: (f(x))^5 + 5 f(x) + \sin(f(x)) = x$ 
Prove that there exists a function $f$ such that
$$\forall x \in \mathbb{R}: (f(x))^5 + 5 f(x) + \sin(f(x)) = x$$

I've been trying to use inverse theorem or using the derivative and can't get anything out of it.
 A: The most elegant and quickest way is that based on the comment by @Alexander Burstein.
Here are two other ways based on your ideas:
Using implicit function theorem (which boils down to above mentioned comment):
$$F(x,y) = y^5+5y+\sin y - x = 0 $$ 
satisfies the assumtions of the implicit function theorem:
$$\partial_yF(x,y) = 5y^4+5+\cos y > 0$$
For any $(x_0,y_0) \in \mathbb{R}$ with $F(x_0,y_0)$ the equation is invertible, which means it defines a function $y=f(x)$ in an open neighbourhood of $(x_0,y_0)$.
Using derivatives:
Just assuming $y$ is differentiable, we get 
$$5y^4y' + 5y'+\cos{(y)}y' - 1 = 0 \Rightarrow y' =\frac{1}{5y^4+5+\cos y}$$
For any $(x_0,y_0) \in \mathbb{R}$ with $F(x_0,y_0)$ we have an initial value problem
$$y'(x)=g(x,y(x))= \frac{1}{5y^4+5+\cos y}, \; y(x_0) = y_0$$
Function $g$ satisfies the assumptions of the Picard-Lindelöf theorem, as $g$ is uniformly Lipschitz continuous in $y$, because $g'$ is bounded.
So, we get the existence of the function looked for.
In both cases you get the unique function $f(x)$ locally.
You may reason for yourself why these local functions represent "snippets" of a global $f(x)$.
