The particular solution of the differential equation x(1 + y²)dx - y(1 + x²)dy = 0, y(0) = 1, is:

$x(1 + y^2)\,dx - y(1 + x^2)\,dy = 0, \quad y(0) = 1$

$x(1 + y^2)\,dx = y(1 + x^2)\,dy$

$\dfrac{x\,dx}{1 + x^2} = \dfrac{y\,dy}{1 + y^2}$

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$\displaystyle\int \frac{x\,dx}{1 + x^2} = \int \frac{y\,dy}{1 + y^2}$

Both integrals are of the form $\displaystyle\int \frac{t\,dt}{1 + t^2}$, where substituting $u = 1 + t^2$ gives $\dfrac{1}{2}\ln|1 + t^2|$.

$\dfrac{1}{2}\ln(1 + x^2) = \dfrac{1}{2}\ln(1 + y^2) + C$

$\ln(1 + x^2) = \ln(1 + y^2) + C_1$

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Applying $y(0) = 1$:

$\ln(1 + 0) = \ln(1 + 1) + C_1$

$0 = \ln 2 + C_1$

$C_1 = -\ln 2$

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$\ln(1 + x^2) = \ln(1 + y^2) - \ln 2$

$\ln(1 + x^2) + \ln 2 = \ln(1 + y^2)$

$\ln\big[2(1 + x^2)\big] = \ln(1 + y^2)$

$2(1 + x^2) = 1 + y^2$

$2 + 2x^2 = 1 + y^2$

$\boxed{y^2 = 2x^2 + 1}$