$\frac{dy}{dx}=\frac{y}{x}+\frac{y^2}{2x^2}$

$\Rightarrow y^{-2}\frac{dy}{dx}-\frac{1}{y}·\frac{1}{x}=\frac{1}{2x^2}$

Put $-\frac{1}{y}=t \Rightarrow \frac{1}{y^2}\frac{dy}{dx}=\frac{dt}{dx}$

$\Rightarrow \frac{dt}{dx}+\left(\frac{1}{x}\right)t=\frac{1}{2x^2}$ 

This is a linear differential equation,

with Integrating Factor : $e^{\int \frac{1}{x}dx}=e^{\ln \left(x\right)}=x$ 

So, solution of the linear differential equation is  $tx=\int \frac{1}{2x^2}·xdx+C$

$\Rightarrow -\frac{x}{y}=\frac{1}{2}\ln \left(x\right)+C$

 The curve passes through $\left(1,2\right)$

 $\Rightarrow -\frac{1}{2}=\frac{1}{2}\ln \left(1\right)+\mathrm{C}\Rightarrow \mathrm{C}=-\frac{1}{2}$ 

Hence, the particular solution to the differential equation is $-\frac{x}{y}=\frac{1}{2}\ln \left(x\right)-\frac{1}{2}$ 

$\Rightarrow \frac{x}{y}=\frac{1-\ln \left(x\right)}{2}\Rightarrow y=\frac{2x}{1-\ln \left(x\right)}$

$\Rightarrow f\left(\frac{1}{2}\right)=\frac{2\times \frac{1}{2}}{1-\ln \left(\frac{1}{2}\right)}=\frac{1}{1+\ln \left(2\right)}=\frac{1}{1+{\log }_e\left(2\right)}$