$\text{$\frac{dy}{dt} = y(1-y)$}$
$\text{$\int \frac{dy}{y(1-y)} =\int dt$}$
$\text{$\int \left ( \frac{1}{y} + \frac{1}{1-y} \right )dy = \int dt$}$
$\text{$\log |y| \; –\; \log |(1-y)| +c = t$ (c is arbitrary constant)}$
$\text{$\log y \; –\; \log (1-y) = t – c$}$
$\text{$\log \frac{y} {(1-y)} = t – c$}$
$\text{$ \frac{y} {1-y} =e^{ t – c}$}$
$\text{$ y= \frac{e^{t-c}}{1+e^{t-c}}= \frac{1}{1+e^{c-t}} = \frac{1}{1+\frac{e^{c}}{e^t}}$}$
As $t \rightarrow \infty, e^t \rightarrow \infty,$ So, $\lim_{t \rightarrow \infty } y(t) = 1$
$\text{But there are 2 }$ trivial $\text{solutions also for the given differential equation i.e. $y(t)=0$ and $y(t)=1$ }$
$\text{which satisfies the given differential equation $y’=y(1-y)$}$
$\text{and so, $\lim_{t \rightarrow \infty } y(t) = \lim_{t \rightarrow \infty } 0 = 0$}$
$\text{So, given statement is False.}$