Existence and uniqueness of solution for the IVP : $y'(x) = [x-\cos(y(x))]^{2/5}, \space y(0) = 0$ Exams Problem :

Consider the IVP :
  $$y'(x) = [x-\cos(y(x))]^{2/5}, \space y(0) = 0$$
  Examine if there exists a solution for the given IVP in an area of $x=0$. Furthermore, examine if conclusions can be made about the uniqueness of the solution.

Attempt :
As I've went over similar posts, you need to always determine how you define the powers/roots in this case, since it's a detail fact that plays a big role.
Let's assume that we consider the real-valued $5th$ root in this particular case. Then, let's consider the function :
$$f(x,y) = [x-\cos(y)]^{2/5}=\sqrt[5]{(x-\cos(y))^{2}}$$
Then, the function $f(x,y)$ will be continuous $\forall (x,y) \in \mathbb R^2$ since it is $(x-\cos(y))^2 \geq 0$.
This means that the function $f(x,y)$ will also be continuous in a domain, such as : 
$$D=\{(x,y) \in \mathbb R^2 :|x|\leq a, |y| \leq b\} \space \text{where} \space a,b > 0$$
Obviously, the Picard/Peano theorems can be applied and then truly, there exists a solution for the IVP around $x=0$ with $y(0)=0$.
My question is regarding the uniqueness though :
One way to determine the uniqueness, is showing that the derivative of $f(x,y)$ with respect to $y$ is bounded or in other words (and approach), Lipschitz Continuous.
Simply then (still considering the real valued root branch case) :
$$\frac{\partial}{\partial y}f(x,y)= \frac{2\sin(y)}{5\sqrt[5]{(x-\cos(y))^3}}$$
But in such case, we see that $f_y(x,y)$ is not continuous for every value, since it's not always $(x-\cos(y))^3 \geq 0$. This means that the derivative $f_y(x,y)$ won't be continuous in the domain $D$ (which is a standard step for the existence) and thus no conclusions can be made about the uniqueness. In another way, the derivative $f_y(x,y)$ is not bounded (as I think), so still no conclusions can be made.
Question : Is the above explanation and reasoning correct though ? Is there a more formal way of showing that $f_y(x,y)$ is not bounded or is my approach mistaken and the derivative could be bounded - defined ?
 A: With intial condition $y(0) = 0$, you do have uniqueness of the local solution. This you can see from a uniqueness criterion we can formulate as follows:

Consider the IVP $\dot y (x)= F(x, y(x))$, where $F : \mathbb R^2 \to \mathbb R$ is Lipschitz on a small open neighborhood $U_0$. Then, the IVP with $y(x_0) = y_0$ is locally well-posed for any $(x_0, y_0) \in U_0$.

Here, local well-posedness refers to the existence of some $T = T(x_0, y_0)$ for which a unique solution $y : [- T, T] \to \mathbb R$ to the IVP exists. We say that $F$ is Lipschitz on $U_0$ if for all $z_1, z_2 \in U_0$, we have $|F(z_1) - F(z_2)| \leq L \| z_1 - z_2\|$, where $L$ is a constant (the Lipschitz constant).
Since $(x_0, y_0) = (0,0)$ is away from the singularity set $\{\cos y = x\}$ where $F$ fails to be $C^1$, your $F$ is Lipschitz on a small neighborhood $U_0$ of the origin. Notice that we are indifferent to properties of $F$ outside of $U_0$.

Uniqueness is typically determined by checking that the Picard map
$$
h \mapsto \Phi[h](x) := h(x_0) + \int_{t = x_0}^x F(t, h(t)) dt
$$
is a contraction mapping on the set of continuous mappings $h : [-T, T] \to \mathbb R, h(x_0) = y_0$ for some $T$ sufficiently small. A fixed point $h = \Phi[h]$ of the Picard map is automatically a differentiable solution to the original IVP. Checking the contraction property typically requires the Lipschitz property of $F$ on $U_0$.
That being said, $F$ is not Lipschitz in neighborhoods of the singularity set $\{ \cos y = x\}$, and so uniqueness can, a priori, fail for some initial values. 
By way of example, let $x_* \in (0,\pi/2)$ be the unique solution to $\cos x_* = x_*$, $x_* \approx .74$. Then, one solution to the IVP $(x_0, y_0) = (x_*, x_*)$ is the constant solution $y(x) \equiv x_*$. 
I would wager that a second, distinct solution can be derived; it would be an interesting exercise to do so.
