Answer
$-0.2$
Work Step by Step
Here, we have $\int \dfrac{dy}{y^2}=\int (2x-2) dx$
and $\dfrac{-1}{y}=x^2-2x+c$
Further, apply the initial conditions.
$-\dfrac{1}{2}=2^2-2(2)+c=4-4+c$
or, $c=2$
Now, we have
$\dfrac{-1}{y}=x^2-2x+c \implies \dfrac{-1}{y}=x^2-2x+2$
Thus, $y=-\dfrac{1}{x^2-2x+2}$
and $y(3)=-\dfrac{1}{3^2-2(3)+2} =-\dfrac{1}{9-6+2} = -0.2$