(c) Find $\frac{d y}{d x}$ where $x+y+\sqrt{x y}=1$.

Step 1 ：Understand the problem: We are asked to find the derivative of y with respect to x, given the equation \(x + y + \sqrt{xy} = 1\).

Step 2 ：Differentiate both sides of the equation with respect to x: \(1 + \frac{dy}{dx} + \frac{1}{2} \cdot \frac{1}{\sqrt{xy}} \cdot (y + x \cdot \frac{dy}{dx}) = 0\).

Step 3 ：Simplify the equation by multiplying through by \(2\sqrt{xy}\) to clear the fraction: \(2\sqrt{xy} + 2\sqrt{xy} \cdot \frac{dy}{dx} + y + x \cdot \frac{dy}{dx} = 0\).

Step 4 ：Collect terms involving \(\frac{dy}{dx}\): \((2\sqrt{xy} + x) \cdot \frac{dy}{dx} = -2\sqrt{xy} - y\).

Step 5 ：Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{-2\sqrt{xy} - y}{2\sqrt{xy} + x}\).

Step 6 ：Check the result by substituting \(\frac{dy}{dx}\) back into the differentiated equation to check if it satisfies the equation. If it does, then the result is correct.

Step 7 ：\(\boxed{\frac{dy}{dx} = \frac{-2\sqrt{xy} - y}{2\sqrt{xy} + x}}\)