Differentiate implicitly to find $\frac{\mathrm{dy}}{\mathrm{dx}}$
\[
8 x y+1=0
\]
\[
\frac{d y}{d x}=\square
\]

Step 1 ：Differentiate both sides of the equation \(8xy + 1 = 0\) with respect to x.

Step 2 ：Using the product rule, the derivative of the left side of the equation is \(d/dx(8xy) = 8x(dy/dx) + 8y\).

Step 3 ：The derivative of 1 is 0.

Step 4 ：So, the derivative of the left side of the equation is \(8x(dy/dx) + 8y\).

Step 5 ：The derivative of the right side of the equation is 0.

Step 6 ：Setting these equal to each other gives \(8x(dy/dx) + 8y = 0\).

Step 7 ：Solve for dy/dx: \(8x(dy/dx) = -8y\).

Step 8 ：Divide both sides by 8x to isolate dy/dx: \((dy/dx) = -8y/8x\).

Step 9 ：Simplify to get the final answer: \(\boxed{(dy/dx) = -y/x}\).