Let $$y = y(x)$$ be the solution of the differential equation $$e^x\sqrt{1-y^2}dx + \left(\frac{y}{x}\right)dy = 0$$, $$y(1) = -1$$. Then the value of $$(y(3))^2$$ is equal to:

Separating variables from $$e^x\sqrt{1-y^2}\,dx + \frac{y}{x}\,dy = 0$$: $$x e^x\,dx = -\frac{y}{\sqrt{1-y^2}}\,dy.$$

Integrating the left side by parts ($$u=x$$, $$dv=e^x dx$$): $$\int x e^x\,dx = e^x(x-1) + C_1.$$

Integrating the right side with the substitution $$u = 1-y^2$$: $$-\int \frac{y}{\sqrt{1-y^2}}\,dy = \sqrt{1-y^2} + C_2.$$

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

Applying the initial condition $$y(1) = -1$$: $$e^1(1-1) = \sqrt{1-(-1)^2} + C \implies 0 = 0 + C$$, so $$C = 0$$.

The solution is $$e^x(x-1) = \sqrt{1-y^2}$$, so $$1 - y^2 = e^{2x}(x-1)^2$$.

At $$x = 3$$: $$(y(3))^2 = 1 - e^{2 \cdot 3}(3-1)^2 = 1 - 4e^6$$.