Find the slope of the tangent line to the curve
\[
x^{2}-3 x y-2 y^{3}=20
\]
at the point $(3,-1)$.

Step 1 ：First, we need to find the derivative of the given equation. The derivative of a function gives us the slope of the tangent line at any point. So, we differentiate the given equation implicitly with respect to \(x\).

Step 2 ：Taking the derivative of \(x^{2}-3 x y-2 y^{3}=20\) with respect to \(x\), we get \(2x - 3y - 3xy' - 6y^{2}y' = 0\).

Step 3 ：Now, we solve this equation for \(y'\), the derivative of \(y\) with respect to \(x\).

Step 4 ：Rearranging the terms, we get \(3xy' + 6y^{2}y' = 2x - 3y\).

Step 5 ：Factoring out \(y'\), we get \(y'(3x + 6y^{2}) = 2x - 3y\).

Step 6 ：So, \(y' = \frac{2x - 3y}{3x + 6y^{2}}\). This is the derivative of \(y\) with respect to \(x\), which gives us the slope of the tangent line at any point \((x, y)\) on the curve.

Step 7 ：Now, we substitute the given point \((3, -1)\) into the derivative to find the slope of the tangent line at this point.

Step 8 ：Substituting \(x = 3\) and \(y = -1\) into \(y' = \frac{2x - 3y}{3x + 6y^{2}}\), we get \(y' = \frac{2(3) - 3(-1)}{3(3) + 6(-1)^{2}} = \frac{6 + 3}{9 + 6} = \frac{9}{15}\).

Step 9 ：Simplifying \(\frac{9}{15}\) gives us \(\frac{3}{5}\).

Step 10 ：So, the slope of the tangent line to the curve at the point \((3, -1)\) is \(\boxed{\frac{3}{5}}\).