$$ \begin{aligned} & \left(1+x^2\right) d y=y(x-y) d x \\\\ & y(0)=1 \cdot y(2 \sqrt{2})=\beta \\\\ & \frac{d y}{d x}=\frac{y x-y^2}{1+x^2} \\\\ & \frac{d y}{d x}+y\left(\frac{-x}{1+x^2}\right)=\left(\frac{-1}{1+x^2}\right) y^2 \\\\ & \frac{1}{y^2} \frac{d y}{d x}+\frac{1}{y}\left(\frac{-x}{1+x^2}\right)=\frac{-1}{1+x^2} \\\\ & \text { put } \frac{1}{y}=t \text { then } \frac{-1}{y^2} \frac{d y}{d x}=\frac{d t}{d x} \end{aligned} $$

$$ \frac{\mathrm{dt}}{\mathrm{dx}}+\mathrm{t} \frac{\mathrm{x}}{1+\mathrm{x}^2}=\frac{1}{1+\mathrm{x}^2} $$

$$ \text { I.F }=\mathrm{e}^{\int \frac{\mathrm{x}}{1+\mathrm{x}^2} \mathrm{dx}}=\mathrm{e}^{\frac{1}{2} \ln \left(1+\mathrm{x}^2\right)}=\sqrt{1+\mathrm{x}^2} $$

$$ t \sqrt{1+x^2}=\int \frac{\sqrt{1+x^2}}{1+x^2} d x $$

$$ \begin{aligned} & \frac{1}{y} \sqrt{1+x^2}=\int \frac{1}{\sqrt{1+x^2}} d x \\\\ & \frac{1}{y} \sqrt{1+x^2}=\ln \left(x+\sqrt{x^2+1}\right)+C \\\\ & \because y(0)=1 \Rightarrow C=1 \\\\ & \frac{1}{y} \sqrt{1+x^2}=\ln \left(x+\sqrt{x^2+1}\right)+1 \\\\ & \text { For } y=2 \sqrt{2} \\\\ & \frac{3}{y}=\ln |2 \sqrt{2}+3|+1 \end{aligned} $$

$$ \begin{gathered} y=\beta=\frac{3}{1+\ln |2 \sqrt{2}+3|} \\\\ \Rightarrow 3 \beta^{-1}=1+\ln |2 \sqrt{2}+3| \end{gathered} $$

Isolating the term $e^{3 \beta^{-1}}$, we get :

$$e^{3 \beta^{-1}} = e^{1+\ln |2 \sqrt{2}+3|}.$$

This can be simplified using the rule $e^{a+b} = e^a e^b$ to :

$$e^{3 \beta^{-1}} = e \cdot e^{\ln |2 \sqrt{2}+3|}.$$

Since $e^{\ln(x)} = x$ for any $x$, this simplifies to :

$$e^{3 \beta^{-1}} = e |2 \sqrt{2}+3|.$$

Using the given value of $\beta$, which is $\frac{3}{1+\ln |2 \sqrt{2}+3|}$, we find :

$$e^{3 \beta^{-1}} = e |2 \sqrt{2}+3| = e (2\sqrt{2} + 3),$$

Since $2\sqrt{2}+3$ is positive and so the absolute value does not affect the result.