Assume the Initial Value Problem:
$$
y'(x)=[x-y(x)]^{\frac23}\equiv f(x,y(x)), \quad y(5)=5
$$
Existence:
Since $f: \mathbb{R^2} \longrightarrow \mathbb{R}$ and $(x_0,y_0)=(5,5) \in \mathbb{R^2}$, Peano's Theorem ensures that there exists a local solution $z: I \longrightarrow \mathbb{R}$, where $I$ is a neighborhood of $x_0=5$.
Uniqueness:
I'm not sure whether we can deduce anything about uniqueness. We could take:
$$
f_y=\frac{\partial}{\partial y} f(x,y)=-\frac{2}{3(x-y)^{\frac13}}, \quad \forall (x,y):x\neq y
$$
Here's the part which confuses me. Does the fact that $f_y$ not bounded for any neighborhood of $x_0=5$ prevents us from proving uniqueness?
Best Answer
As you have already seen, the function $f$ is not locally Lipschitz continuous around $y=x$, hence you cannot deduce uniqueness using the standard Cauchy-Lipschitz theory.
Nevertheless, the Cauchy problem admits a unique solution. You can get an intuition of this fact observing that any solution hits the "bad" set $\{(x,y):\ x=y\}$ only for $x=5$, since $y'(5) = 0$.
For a formal proof, you can observe that $y(x)$ is a solution of your Cauchy problem if and only if the function $z(x) := x - y(x)$ is a solution to $$ z' = 1-z^{2/3}, \quad, z(5) = 0, $$ and this Cauchy problem admits unique solution.