Separate the variables:
[tex]y' = \dfrac{dy}{dx} = (3+y)(1-y) \implies \dfrac{dy}{(3+y)(1-y)} = dx[/tex]
On the left, take the partial fraction decomposition:
[tex]\dfrac{1}{(3+y)(1-y)} = \dfrac a{3+y} + \dfrac b{1-y}[/tex]
[tex]\dfrac{1}{(3+y)(1-y)} = \dfrac{a(1-y) + b(3+y)}{(3+y)(1-y)}[/tex]
[tex]\dfrac{1}{(3+y)(1-y)} = \dfrac{(b-a)y+a-3b}{(3+y)(1-y)}[/tex]
[tex]\implies (b-a)y +a-3b = 1[/tex]
[tex]\implies \begin{cases}b-a=0\\a-3b=1\end{cases} \implies a=b=-\dfrac12[/tex]
[tex]\implies \dfrac1{(3+y)(1-y)} = -\dfrac12 \left(\dfrac1{3+y} + \dfrac1{1-y}\right)[/tex]
Now integrate both sides:
[tex]\displaystyle \int -\frac12 \left(\frac1{3+y} + \frac1{1-y}\right) \, dy = \int dx[/tex]
[tex]-\dfrac12 \left(\ln|3+y| + \ln|1-y|\right) = x + C[/tex]
Normally, at this point you would solve for C using the initial condition. However, in this case we have y(0) = 1, and ln|1 - 1| = \ln(0) is undefined.
Notice that if we let y be a constant function, then y' = 0.
Knowing that y(0) must be equal to 1, let's take our solution to be y(x) = 1. Then y'(x) = 0, and on the right side we have 1 - y = 1 - 1 = 0 as well.
So, the solution to this equation is y(x) = 1.
