Euler method fails to approximate the exact solution $y(x)=(2x/3)^{3/2}$ How can I show that the Euler method fails to approximate the exact solution 
$$y(x)=(2x/3)^{3/2}$$ 
to the IVP 
$$y'=y^\frac{1}{3}$$
$$y(0)=0$$ 
Here we have $f(t,y)= y^\frac{1}{3}$,  $y_0=0$ and so $f(t_0,y_0)=f(0,0)=0$ and 
$$y_{n+1}=y_n +h f(t_n,y_n)$$
Thus
$$y_1=0 \\
y_2=0 \\
\vdots\\
y_n=0$$
So, I can't understand why it fails. Could you help me?

Note that the general solution,
$$ y(x) = \left(\frac{2x+C}{3}\right)^{3/2}$$
Can be achieved by integration:
$$\int \frac{dy }{y^{1/3}}=\int dx$$
This assumes that $y\neq 0$, which is not necessarily true. You therefore have two solutions at $x=0$, and you can't force the Euler method to "choose" the right one.
The reason you don't have a unique solution at $x=0$ is that $d(y^{1/3})/dx$ does not exist on any open interval containing $x=0$. Choosing an IV point with $x> 0$ will guarantee local uniqueness.
