Let y = y (x) be the solution of the differential equation e ^ x 1...

We have,

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

$⇒ {e}^{x}\sqrt{1 - {y}^{2}}dx = \left(\dfrac{- y}{x}\right)dy$

$⇒ ∫ \dfrac{- y}{\sqrt{1 - {y}^{2}}}dy = ∫ x{e}^{x}dx$

$⇒ \dfrac{1}{2} ∫ \dfrac{d\left(1 - {y}^{2}\right)}{\sqrt{1 - {y}^{2}}}dy = ∫ x{e}^{x}dx$

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

Now, at $x = 1, y = - 1$, then

$⇒ 0 = 0 + c ⇒ c = 0$

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

At $x = 3$, then

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

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

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