Find the equation of the tangent line at the given point on the curve.
$$3y^2-\sqrt{x}=8,(16,2)$$
$$y=$$

Step 1 ：First, we need to find the derivative of the given function. The derivative of $3y^2-\sqrt{x}=8$ with respect to $x$ is $\frac{dy}{dx}=\frac{1}{6y\sqrt{x}}$.

Step 2 ：Substitute the given point $(16,2)$ into the derivative to find the slope of the tangent line. So, $\frac{dy}{dx}=\frac{1}{6\ast 2\ast \sqrt{16}}=\frac{1}{24}$.

Step 3 ：The equation of the tangent line is $y-y_1=m(x-x_1)$, where $m$ is the slope of the line and $(x_1,y_1)$ is a point on the line.

Step 4 ：Substitute $m=\frac{1}{24}$, $x_1=16$, and $y_1=2$ into the equation of the tangent line to get $y-2=\frac{1}{24}(x-16)$.

Step 5 ：Rearrange the equation to get $y=\frac{1}{24}x+\frac{2}{3}$.

Step 6 ：Check the equation by substituting the point $(16,2)$ into it. We get $2=\frac{1}{24}\ast 16+\frac{2}{3}=2$, which verifies that the equation is correct.

Step 7 ：So, the equation of the tangent line at the point $(16,2)$ on the curve $3y^2-\sqrt{x}=8$ is $\boxed{{\displaystyle y=\frac{1}{24}x+\frac{2}{3}}}$.