Use implicit differentiation to find $dy/dx$. Then find the slope of the curve at the given point.
$$5xy-8x+y=40;(-4,-\frac{8}{19})$$
$$\frac{dy}{dx}=$$

Step 1 ：First, we need to differentiate both sides of the equation with respect to $x$. The left side of the equation is a product of two functions, $5x$ and $y$, so we need to use the product rule. The derivative of $5xy$ with respect to $x$ is $5y+5x\frac{dy}{dx}$. The derivative of $-8x$ with respect to $x$ is $-8$. The derivative of $y$ with respect to $x$ is $\frac{dy}{dx}$. So, the derivative of the left side of the equation is $5y+5x\frac{dy}{dx}-8+\frac{dy}{dx}$. The right side of the equation is a constant, so its derivative is 0.

Step 2 ：Setting the derivative equal to 0 gives us the equation $5y+5x\frac{dy}{dx}-8+\frac{dy}{dx}=0$. We can simplify this to $5x\frac{dy}{dx}+\frac{dy}{dx}=8-5y$.

Step 3 ：Factoring out $\frac{dy}{dx}$ gives us $\frac{dy}{dx}(5x+1)=8-5y$.

Step 4 ：Solving for $\frac{dy}{dx}$ gives us $\frac{dy}{dx}=\frac{8-5y}{5x+1}$.

Step 5 ：Substituting the given point $(-4,-\frac{8}{19})$ into the equation gives us $\frac{dy}{dx}=\frac{8-5(-\frac{8}{19})}{5(-4)+1}$.

Step 6 ：Simplifying the right side of the equation gives us $\frac{dy}{dx}=\frac{8+\frac{40}{19}}{-20+1}=\frac{8\times 19+40}{-19}=\frac{112}{-19}$.

Step 7 ：So, the slope of the curve at the given point is $\boxed{{\displaystyle -\frac{112}{19}}}$.