Let a smooth curve $y=f\left(x\right)$ be such that the slope of the tangent at any point $\left(x,y\right)$ on it is directly proportional to $\left(\frac{-y}{x}\right)$. If the curve passes through the points $\left(1,2\right)$ and $\left(8,1\right)$, then $\left|y\left(\frac{1}{8}\right)\right|$ is equal to

Given, $\frac{dy}{dx}\propto \frac{-y}{x}$

$\Rightarrow \frac{dy}{dx}=\frac{-Ky}{x}$ (where $K$ is proportionality constant)

Now integrating both side, 

$\Rightarrow \int \frac{dy}{y}=-K\int \frac{dx}{x}$

$\Rightarrow \ln \left|y\right|=-K\ln \left|x\right|+C$

If the above equation satisfy $\left(1,2\right)$

$\Rightarrow \ln 2=-K\times 0+C\Rightarrow C=\ln 2$

So, $\ln \left|y\right|=-K\ln \left|x\right|+\ln 2$

Now it also passes through $\left(8,1\right)$

$\Rightarrow \ln 1=-K\ln 8+\ln 2\Rightarrow K=\frac{1}{3}$

So, equation becomes $\ln \left|y\right|=-\frac{1}{3}\ln \left|x\right|+\ln 2$

So, at $x=\frac{1}{8}$

$\Rightarrow \ln \left|y\right|=-\frac{1}{3}\ln \left(\frac{1}{8}\right)+\ln 2=2\ln 2$

$\Rightarrow \left|y\right|=4$