Existence and uniqueness of solution of : $y'(x) = (x-y(x))^{4/5}, \space y(3)=3$ 
Study the Existence and Uniqueness of the IVP :
  $$y'(x) = (x-y(x))^{4/5}, \space y(3)=3$$

Regarding existence, I'd first assume $f$ the function such : 
$$f(x,y)=(x-y)^{4/5}$$
We observe that $f(x,y)$ is not continuous in $
\mathbb R^2$ since we need $x-y\geq0$.
Letting $D$ be a domain such : 
$$D=\{(x,y)\in \mathbb R^2: |x-3| \leq ε_1, \space |y-3|\leq ε_2 \}$$
we can say that there exists $ε_1,ε_2 \in \mathbb R$ such that $f$ is continuous in $D$, which means that the IVP has a solution in $D$.
Question : Is the above statement correct ? Shall I go into more details explaining why $f$ can be continuous in such a domain ?
Finally, regarding uniqueness, I know that showing that $f(x,y)=(x-y)^{4/5}$ is a Lipschitz function is enough. : 
$$|f(x,y_2) -f(x,y_1)|=|(x-y_2)^{4/5}-(x-y_1)^{4/5}|$$
The function $g(y)=(x-y)^{4/5}$ is continuous in the interval $[y_1,y_2]$ with $y_1,y_2 \in D_f$ and $g$ is also differentiable in $(y_1,y_2)$ , which means that we can apply the Mean Value Theorem and yield :
$$g'(ξ)=\frac{g(y_2)-g(y_1)}{y_2-y_1}\Rightarrow \bigg|\frac{4}{5(x-ξ)^{1/5}}\bigg||y_2-y_1|=|(x-y_2)^{4/5}-(x-y_1)^{4/5}|$$
Here, we can see that there isn't a bound for the expression $\bigg|\frac{4}{5(x-ξ)^{1/5}}\bigg|$, thus meaning that the solution is not unique.
 A: The function $f$ is not defined in your set $D$ since $D$ contains points where $x<y$. So your statement is not correct.
Also for uniqueness, the theorem says that if $f$ is Lipschitz in $y$, then the solution is unique, but the converse is not true. So you cannot say that since $f$ is not Lipschitz, there is no uniqueness. 
It is true that one could define $t^{4/5}=(t^4)^{1/5}$ but then you lose important properties like $t^{ab}=(t^a)^b$. In general it is better not to define $t^a$ if $a$ is a rational and $t<0$.
A way out in this case would be to extend $f$ to be zero in the region $x<y$. So now the problem $y'(x)=f(x,y)$, $y(3)=3$ has a solution since $f$ is continuous in $\mathbb R^2$. Then you can follow Riegel's proof to get existence and show that the solution that you get stays in the region $x-y\ge 0$ so it is a solution of your original equation. 
A: The function $f(x,y) := (x-y)^{4/5}$ is continuous is $\mathbb{R}^2$, being the composition of the continuous functions $\phi(t) := t^{4/5} = \sqrt[5]{t^4}$, $t\in\mathbb{R}$ and $g(x,y) := x-y$, $(x,y)\in\mathbb{R}^2$.
By Peano's existence theorem, the Cauchy problems associated with $f$ do have solutions.
The function $f$ is not (locally) Lipschitz continuous, but this is only a sufficient condition in order to get uniqueness of solution of Cauchy problems.
It is easy to see that $y(x)$ is a solution of $y'=f(x,y)$ if and only if 
$z(x) := y(x) - x$ is a solution of $z' = h(x,z)$, with $h(x,z) := z^{4/5} - 1$.
In particular, $y(x)$ is a solution of the given Cauchy problem if and only if $z(x) := y(x) - x$ is a solution of the Cauchy problem
$$
\begin{cases}
z' = z^{4/5} - 1,\\
z(3) = 0.
\end{cases}
$$
This Cauchy problem admits, locally, a unique solution, implicitly defined by
$$
\int_0^z \frac{1}{s^{4/5}-1}\, ds = x-3.
$$
It can be proved (but this requires a simple qualitative study) that this unique local solution can be prolonged to a unique global solution.
