Answer
The solution of initial value problem is given by
$$
tan ^{-1}(y)=2x+x^{2} .
$$
$y$ will be minimum at $x=-1$
Work Step by Step
$$
y^{'}=\frac{dy}{dx}=2(1+x)(1+y^{2}),\quad \quad y(0)=0
$$
the differential equation can be written as
$$
\frac{dy}{(1+y^{2})}=2(1+x)dx,\quad \quad y(0)=0
$$
integrating the left side with respect to $y$ and the right side with respect to $x$ gives
$$
tan ^{-1}(y)=2x+x^{2}+c \quad \quad (*)
$$
where c is an arbitrary constant.
To determine the solution satisfying the prescribed initial condition, we substitute $x=0$ and $y=0$ in eq. (*)
Hence the solution of initial value problem is given by
$$
tan ^{-1}(y)=2x+x^{2} .
$$
So, $y$ will be minimum where $2x+x^{2}$ is maximum
so, it is maximum if $\frac{d}{dx}(2x+x^{2}) =0.$ at $x=-1$
