Step 1: Square first:
$(\sqrt{y+x} + \sqrt{y-x})^2 = c^2$ gives $2y + 2\sqrt{y^2-x^2} = c^2$, so $y + \sqrt{y^2 - x^2} = \frac{c^2}{2}$, a constant.
Step 2: Differentiate this simpler form:
$y' + \frac{y y' - x}{\sqrt{y^2-x^2}} = 0$, so $y'\left(1 + \frac{y}{\sqrt{y^2-x^2}}\right) = \frac{x}{\sqrt{y^2-x^2}}$, which gives $y' = \frac{x}{\sqrt{y^2-x^2} + y}$.
Step 3: Rationalise:
Multiply by $y - \sqrt{y^2-x^2}$ over itself: $y' = \frac{x(y - \sqrt{y^2-x^2})}{x^2} = \frac yx - \sqrt{\frac{y^2}{x^2}-1}$. Hence $K = y/x$.
Final Answer:
K is y over x, option (D).
\[ \boxed{\frac{y}{x}} \]