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

Step 1 ：Given the function $y=\frac{e^{8x}}{x}$ and the point $(\frac{1}{8},8e)$, 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 $y^{\prime }=8\frac{e^{8x}}{x}-\frac{e^{8x}}{x^2}$.

Step 5 ：Evaluating the derivative at the given point $(\frac{1}{8},8e)$, 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 $y=\frac{e^{8x}}{x}$ at the point $(\frac{1}{8},8e)$ is $\boxed{{\displaystyle -8e+\frac{e^{8x}}{x}=0}}$.