If $x\sqrt{1 + y} + y\sqrt{1 + x} = 0$, where $|x| < 1, |y| < 1$ and $x ≠ y$, then

Given $x\sqrt{1 + y} + y\sqrt{1 + x} = 0$ where $|x| < 1, |y| < 1$ and $x ≠ y$.

$x\sqrt{1 + y} = -y\sqrt{1 + x}$

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

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

$x^2 - y^2 = y^2x - x^2y$

$x^2 - y^2 = -xy(x - y)$

$(x - y)(x + y) = -xy(x - y)$

Since $x ≠ y$, dividing both sides by $(x - y)$:

$x + y = -xy$

$x + y + xy = 0$

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Differentiating $x + y + xy = 0$ with respect to $x$:

$\dfrac{d}{dx}(x) + \dfrac{d}{dx}(y) + \dfrac{d}{dx}(xy) = 0$

$1 + \dfrac{dy}{dx} + y + x\dfrac{dy}{dx} = 0$

$1 + y + \dfrac{dy}{dx}(1 + x) = 0$

$\dfrac{dy}{dx} = -\dfrac{1 + y}{1 + x}$

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Checking if $(1 + x)^2 \dfrac{dy}{dx} + 1 = 0$:

$(1 + x)^2 \cdot \left(-\dfrac{1 + y}{1 + x}\right) + 1$

$= -(1 + x)(1 + y) + 1$

$= -(1 + x + y + xy) + 1$

Since $x + y + xy = 0$:

$= -(1 + 0) + 1$

$= -1 + 1$

$= 0$

Therefore, $(1 + x)^2 \dfrac{dy}{dx} + 1 = 0$ is correct.