Find the slope of the tangent line to the curve
$$x^2-3xy-2y^3=20$$
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 $x^2-3xy-2y^3=20$ with respect to $x$, we get $2x-3y-3x{y}^{\prime }-6y^2{y}^{\prime }=0$.

Step 3 ：Now, we solve this equation for $y^{\prime }$, the derivative of $y$ with respect to $x$.

Step 4 ：Rearranging the terms, we get $3x{y}^{\prime }+6y^2{y}^{\prime }=2x-3y$.

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

Step 6 ：So, $y^{\prime }=\frac{2x-3y}{3x+6y^2}$. 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 $y^{\prime }=\frac{2x-3y}{3x+6y^2}$, we get $y^{\prime }=\frac{2(3)-3(-1)}{3(3)+6(-1{)}^2}=\frac{6+3}{9+6}=\frac{9}{15}$.

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 $\boxed{{\displaystyle \frac{3}{5}}}$.