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

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, \(\frac{dy}{dx} = \frac{1}{6*2*\sqrt{16}} = \frac{1}{24}\).

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 \(2 = \frac{1}{24}*16 + \frac{2}{3} = 2\), 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 \(\boxed{y = \frac{1}{24}x + \frac{2}{3}}\).