Find an equation of the tangent line to the given curve at the specified point.
\[
y=\frac{e^{8 x}}{x},\left(\frac{1}{8}, 8 e\right)
\]
\[
y=
\]

Step 1 ：Given the function \(y=\frac{e^{8 x}}{x}\) and the point \(\left(\frac{1}{8}, 8 e\right)\), we are asked to find the equation of the tangent line to the curve at the specified point.

Step 2 ：The equation of a tangent line to a curve at a given point can be found using the formula \(y = mx + b\), where \(m\) is the slope of the tangent line and \(b\) is the y-intercept.

Step 3 ：The slope of the tangent line is the derivative of the function at the given point. So, the first step is to find the derivative of the function.

Step 4 ：The derivative of the function \(y = \frac{e^{8x}}{x}\) is \(y' = 8\frac{e^{8x}}{x} - \frac{e^{8x}}{x^2}\).

Step 5 ：Evaluating the derivative at the given point \(\left(\frac{1}{8}, 8 e\right)\), we find that the slope of the tangent line is 0.

Step 6 ：Now, we can use the point-slope form of the equation of a line to find the equation of the tangent line. The point-slope form is \(y - y1 = m(x - x1)\), where \((x1, y1)\) is the given point and \(m\) is the slope.

Step 7 ：Substituting the given point and the slope into the point-slope form, we get the equation of the tangent line as \(-8e + \frac{e^{8x}}{x} = 0\).

Step 8 ：Final Answer: The equation of the tangent line to the curve \(y=\frac{e^{8 x}}{x}\) at the point \(\left(\frac{1}{8}, 8 e\right)\) is \(\boxed{-8e + \frac{e^{8x}}{x} = 0}\).