Find $\frac{d y}{d x}$ \[ y=\cot x-6 \sqrt{x}+\frac{8}{e^{x}} \]

Step 1 ：First, we need to find the derivative of each term in the equation separately.

Step 2 ：For the first term, \(y = \cot x\), the derivative is \(-\csc^2 x\).

Step 3 ：For the second term, \(y = -6 \sqrt{x}\), the derivative is \(-3x^{-\frac{1}{2}}\).

Step 4 ：For the third term, \(y = \frac{8}{e^x}\), the derivative is \(-\frac{8}{e^x}\).

Step 5 ：Adding these derivatives together, we get \(\frac{d y}{d x} = -\csc^2 x -3x^{-\frac{1}{2}} -\frac{8}{e^x}\).

Step 6 ：Thus, the derivative of the given function is \(\boxed{-\csc^2 x -3x^{-\frac{1}{2}} -\frac{8}{e^x}}\).