We have,

$e^x\sqrt{1-y^2} dx+\left(\frac{y}{x}\right)dy=0, y\left(1\right)=-1$

$\Rightarrow e^x\sqrt{1-y^2} \mathrm{dx}=\left(\frac{-y}{x}\right)dy$

$\Rightarrow \int \frac{-y}{\sqrt{1-y^2}} dy=\int x{e}^{x}dx$

$\Rightarrow \frac{1}{2}\int \frac{d\left(1-y^2\right)}{\sqrt{1-y^2}} dy=\int x{e}^{x}dx$

$\Rightarrow \sqrt{1-y^2}=e^x\left(x-1\right)+c$

Now, at $x=1,y=-1$, then

$\Rightarrow 0=0+c\Rightarrow c=0$

$\therefore \sqrt{1-y^2}=e^x\left(x-1\right)$

At $x=3$, then

$\sqrt{1-{\left(y\left(3\right)\right)}^2}=2e^3$

$\Rightarrow 1-{\left(y\left(3\right)\right)}^2=4e^6$

$\Rightarrow {\left(y\left(3\right)\right)}^2=1-4e^6$