All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Differentiating the given equation by $x$,

$\Rightarrow 2x - 2y \frac{dy}{dx} + 2x \frac{dy}{dx} + 2y = 0$

$(x-y)\frac{dy}{dx} = -y-x$

$\frac{dy}{dx} = \frac{x+y}{y-x}$

(ii)
Tangent parallel to x-axis $\Rightarrow \frac{dy}{dx} = 0$

Then, $y= -x$

$\Rightarrow x^2 - (-x)^2 + 2x (-x)+4=0$

$x^2 = 2$

$x = \pm \sqrt{2}$

$y = -\sqrt{2}, ~\sqrt{2}$

Therefore, coordinates are $(\sqrt{2}, -\sqrt{2})$ and $(-\sqrt{2}, \sqrt{2})$.