Answer
$y=\dfrac{x}{x+1}$
Work Step by Step
We are given that $\dfrac{dy}{dx}=\dfrac{y^2}{x^2}$
We will separate the variables to obtain:
$\dfrac{\ dy}{y^2}=\dfrac{\ dx}{x^2}$
Integrate to obtain:
$\int y^{-2} \ dy=\int x^{-2} \ dx$
This implies that $\dfrac{-1}{y}=\dfrac{-1}{x}+C$
After applying the initial conditions, $y=\dfrac{1}{2}$ when $x=1$, we get $C=-1$
Therefore, we have: $\dfrac{-1}{y}=\dfrac{-1}{x}-1 \implies y=\dfrac{x}{x+1}$
