The domain of a solution and stability of solutions of a differential equation.

A. Assuming $y \neq 0$ and $y \neq 1$, we have
\[\frac{dy}{y-y^{3/5}}=dt   \Rightarrow  \frac{dy}{y^{3/5}(y^{2/5}-1)}=dt.\]
Hence 

\[t=\int \frac{dy}{y^{3/5}(y^{2/5}-1)}=\frac{5}{2}\int\frac{du}{u}=\frac{5}{2}\ln(u)+c=\frac{5}{2}\ln(y^{2/5}-1)+c,\]
where $u=y^{2/5}-1$ with $du=\frac{2}{5}y^{-3/5}=\frac{2}{5y^{3/5}}.$
Hence
\[t=\frac{5}{2}\ln(y^{2/5}-1)+c,\]
Since $y(0)=y_0$, we have
\[0=\frac{5}{2}\ln(y_0^{2/5}-1)+c \Rightarrow c=-\frac{5}{2}\ln(y_0^{2/5}-1),\]
and 
\[t=\frac{5}{2}\ln(y^{2/5}-1)-\frac{5}{2}\ln(y_0^{2/5}-1).\]
Combine the logarithmic terms:
\[ t = \frac{5}{2} \ln \left( \frac{y^{2/5} - 1}{y_0^{2/5} - 1} \right) \]

Divide both sides by $\frac{5}{2}$:
\[ \frac{2}{5} t = \ln \left( \frac{y^{2/5} - 1}{y_0^{2/5} - 1} \right) \]

Exponentiate both sides:
\[ e^{\frac{2}{5} t} = \frac{y^{2/5} - 1}{y_0^{2/5} - 1} \]

Multiply both sides by $ y_0^{2/5} - 1 $:
\[ e^{\frac{2}{5} t} (y_0^{2/5} - 1) = y^{2/5} - 1 \]

Add 1 to both sides:
\[ y^{2/5} = e^{\frac{2}{5} t} (y_0^{2/5} - 1) + 1 \]

Raise both sides to the power of $\frac{5}{2}$:
\[ y = \left( e^{\frac{2}{5} t} (y_0^{2/5} - 1) + 1 \right)^{5/2} \]

Thus, the solution for $ y $ is:
\[ y = \left( e^{\frac{2}{5} t} (y_0^{2/5} - 1) + 1 \right)^{5/2}= \sqrt{\left( e^{\frac{2}{5} t} (y_0^{2/5} - 1) + 1 \right)^{5}}.\]

B. Note that if $y_0^{2/5} - 1<0$, then the expression under the square root becomes negative for large $t$ and the solution does not exists for all $t \in \mathbb{R}$. So the answer for part B is $y_0 \in (-1,1)$. The maximum domain in this case is ontained as follows
\[e^{\frac{2}{5} t_{\max}} (y_0^{2/5} - 1) + 1=0.\]
Solving this equation for $t_{\max}$ we get 
\[ t_{\max} = \frac{5}{2} \ln \left( \frac{-1}{y_0^{2/5} - 1} \right). \]
The maximum domain that the solution exists is 
\[(-\infty,\frac{5}{2} \ln \left( \frac{-1}{y_0^{2/5} - 1}\right) ). \]
C. For $y_0=0$ we have two solutions: 1) $y\equiv 0$ and 2) 

\[ y = \sqrt{\left(- e^{\frac{2}{5} t} + 1 \right)^{5}}.\]

D. To find the fixed solutions we set
\[y-y^{3/5}=0 \Rightarrow y^{3/5}(y^{2/5}-1)=0.\]
Hence $y_0=0$ and $y_0=\pm1$ are setpoints. To determine stability we differentiate $f(y)=y-y^{3/5}$: 
\[f'(y)=1-\frac{3}{5}y^{2/5}.\]

Since $f'(0)=1>0$, the fixed solution $y=0$ is unstable. 

Since $f'(\pm1)=1>0$, the fixed solution $y=\pm 1$ is unstable.