Find $\frac{d y}{d x}$ by implicit differentiation.
\[
\begin{array}{r}
\sin (x)+\cos (y)=9 x-2 y \\
\frac{d y}{d x}=\square
\end{array}
\]

Step 1 ：Differentiate both sides of the equation with respect to x: \(\frac{d}{dx}(\sin(x) + \cos(y)) = \frac{d}{dx}(9x - 2y)\)

Step 2 ：Apply the chain rule to the left side: \(\cos(x) + \frac{d}{dy}(\cos(y)) \cdot \frac{dy}{dx} = 9 - 2 \cdot \frac{dy}{dx}\)

Step 3 ：Differentiate \(\cos(y)\) with respect to y: \(\frac{d}{dy}(\cos(y)) = -\sin(y)\)

Step 4 ：Substitute the result back into the equation: \(\cos(x) - \sin(y) \cdot \frac{dy}{dx} = 9 - 2 \cdot \frac{dy}{dx}\)

Step 5 ：Isolate \(\frac{dy}{dx}\): \(\frac{dy}{dx} - 2 \cdot \frac{dy}{dx} = \cos(x) - \sin(y)\)

Step 6 ：Combine like terms: \(-\frac{dy}{dx} = \cos(x) - \sin(y)\)

Step 7 ：Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = -(\cos(x) - \sin(y))\)

Step 8 ：Final Answer: \(\boxed{\sin(y) - \cos(x)}\)