The particular solution of the differential equation $e^x\sqrt{1-y^2}dx + \frac{y}{x}dy = 0$, given that $y = 1$, when $x = 0$ is:

The differential equation is $e^x\sqrt{1-y^2}dx + \frac{y}{x}dy = 0$ with initial condition $y = 1$ when $x = 0$.

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

$x e^x\sqrt{1-y^2}dx = -y \, dy$

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

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

For the left side, using integration by parts with $u = x$ and $dv = e^x dx$:

$\int x e^x dx = xe^x - e^x$

$= e^x(x-1)$

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For the right side, let $u = 1-y^2$, then $du = -2y \, dy$:

$\int -\frac{y}{\sqrt{1-y^2}}dy = \frac{1}{2}\int u^{-1/2}du$

$= \sqrt{u}$

$= \sqrt{1-y^2}$

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Combining both sides:

$e^x(x-1) = \sqrt{1-y^2} + C$

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Applying the initial condition $y = 1$ when $x = 0$:

$e^0(0-1) = \sqrt{1-1^2} + C$

$-1 = 0 + C$

$C = -1$

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Substituting $C = -1$:

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

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

Therefore, the particular solution is $\sqrt{1-y^2} = (x-1)e^x + 1$.