Solution:
Given curve is
\( (1+x^{2})\,dy=y(x-y)\,dx \) and \( y=y(x) \) Where \( y>0 \) is a solution of it \( \quad\therefore \) we have
\( \dfrac{dy}{dx}=\left(\dfrac{x}{1+x^{2}}\right)y-\dfrac{y^{2}}{1+x^{2}} \)
\( \dfrac{dy}{dx}-\left(\dfrac{x}{1+x^{2}}\right)y=\dfrac{-y^{2}}{1+x^{2}} \)
\( \dfrac{-1}{y^{2}}\dfrac{dy}{dx}+\dfrac{1}{y}\left(\dfrac{x}{1+x^{2}}\right)=\dfrac{1}{1+x^{2}} \)
Put \( \dfrac{1}{y}=z \), s.t. \( \dfrac{-1}{y^{2}}\dfrac{dy}{dx}=\dfrac{dz}{dx} \)
We get \( \dfrac{dz}{dx}+z\left(\dfrac{x}{1+x^{2}}\right)=\dfrac{1}{1+x^{2}} \)
Hence, solution is given by
\( z.\,e^{\displaystyle\int \frac{x}{1+x^{2}}dx}=\displaystyle\int \dfrac{1}{1+x^{2}}\cdot e^{\displaystyle\int \frac{x}{1+x^{2}}dx}dx+C \)
Since \(\displaystyle \int \dfrac{x}{1+x^{2}}dx=\log_{e}\sqrt{1+x^{2}} \)
We get
\( z.\,e^{\log_{e}\sqrt{1+x^{2}}}=\displaystyle\int \dfrac{1}{1+x^{2}}\,e^{\log_{e}\sqrt{1+x^{2}}}dx+C \)

\( z\sqrt{1+x^{2}}=\displaystyle\int \dfrac{\sqrt{1+x^{2}}}{1+x^{2}}dx+C \quad \) (using \( e^{\log_{e}x}=x \))
\( z\sqrt{1+x^{2}}=\displaystyle\int \dfrac{dx}{\sqrt{1+x^{2}}}+C \)
\( z\sqrt{1+x^{2}}=\log|x+\sqrt{1+x^{2}}|+C \Rightarrow z(x)=\dfrac{1}{\sqrt{1+x^{2}}}\left(\log|x+\sqrt{1+x^{2}}|+C\right) \)
\( \Rightarrow \dfrac{1}{y(x)}=\dfrac{1}{\sqrt{1+x^{2}}}\left(\log|x+\sqrt{1+x^{2}}|+C\right)\Rightarrow y(x)=\dfrac{\sqrt{1+x^{2}}}{\left(\log|x+\sqrt{1+x^{2}}|+C\right)} \)
As \( y(0)=1 \quad\therefore 1=\dfrac{\sqrt{1+0}}{(\log1+C)}\Rightarrow C=1 \therefore\quad y(x)=\dfrac{\sqrt{1+x^{2}}}{\log|x+\sqrt{1+x^{2}}|+1} \)
So now, \( y(2\sqrt{2})=\beta \Rightarrow \dfrac{\sqrt{1+8}}{(\log|2\sqrt{2}+\sqrt{1+8}|+1)}=\beta \)

\( \dfrac{3}{(\log|2\sqrt{2}+3|+1)}=\beta \Rightarrow \beta=\dfrac{3}{\log_{e}e(2\sqrt{2}+3)} \) \( \Rightarrow \dfrac{3}{\beta}=\log_{e}e(2\sqrt{2}+3) \)
Hence, \( e^{3\beta^{-1}}=e(2\sqrt{2}+3) \).

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