Use implicit differentiation to find the equation of the tangent line to the curve $x y^{3}+x y=12$ at the point $(6,1)$. The equation of this tangent line can be written in the form $y=m x+b$ where $m$ is: and where $b$ is:

Step 1 ：First, we need to differentiate the given equation implicitly with respect to \(x\). The given equation is \(x y^{3}+x y=12\).

Step 2 ：Applying the product rule to \(x y^{3}\) and \(x y\), we get \(y^{3} + 3xy^{2}y' + y + xy' = 0\).

Step 3 ：Rearranging the equation to solve for \(y'\), we get \(y' = -\frac{y^{3} + y}{3xy^{2} + x}\).

Step 4 ：Substitute \((6,1)\) into the equation to find the slope \(m\) of the tangent line. We get \(m = -\frac{1^{3} + 1}{3*6*1^{2} + 6} = -\frac{1}{24}\).

Step 5 ：The equation of the tangent line is of the form \(y = mx + b\). We already know that \(m = -\frac{1}{24}\).

Step 6 ：Substitute \((6,1)\) into the equation \(y = mx + b\) to solve for \(b\). We get \(1 = -\frac{1}{24}*6 + b\), which simplifies to \(b = 1 + \frac{1}{4} = \frac{5}{4}\).

Step 7 ：So, the equation of the tangent line is \(y = -\frac{1}{24}x + \frac{5}{4}\).

Step 8 ：Finally, the value of \(b + m\) is equal to \(-\frac{1}{24} + \frac{5}{4} = \frac{19}{24}\).

Step 9 ：\(\boxed{\frac{19}{24}}\) is the final answer.