The point on the curve x^2 4 + y^2 9 = 1 at which the tangent to the curve is parallel to the x-axis is

Given $x\sqrt{1 + y} + y\sqrt{1 + x} = 0$x1+y​+y1+x​=0 where $|x| < 1, |y| < 1$∣x∣<1,∣y∣<1 and $x ≠ y$x=y.

$$x\sqrt{1 + y} = -y\sqrt{1 + x}$$x1+y​=−y1+x​

$$x^2(1 + y) = y^2(1 + x)$$x2(1+y)=y2(1+x)

$$x^2 + x^2y = y^2 + y^2x$$x2+x2y=y2+y2x

$$x^2 - y^2 = y^2x - x^2y$$x2−y2=y2x−x2y

$$x^2 - y^2 = -xy(x - y)$$x2−y2=−xy(x−y)

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

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

$$x + y = -xy$$x+y=−xy

$$x + y + xy = 0$$x+y+xy=0

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

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

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

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

$$\dfrac{dy}{dx} = -\dfrac{1 + y}{1 + x}$$dxdy​=−1+x1+y​

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

$$(1 + x)^2 \cdot \left(-\dfrac{1 + y}{1 + x}\right) + 1$$(1+x)2⋅(−1+x1+y​)+1

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

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

Since $x + y + xy = 0$x+y+xy=0:

$$= -(1 + 0) + 1$$=−(1+0)+1

$$= -1 + 1$$=−1+1

$$= 0$$=0

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