Find (a) the slope of the curve at the given point $P$, and (b) an equation of the tangent line at $P$.
$$y=-2-4x^2,P(3,-38)$$

Step 1 ：Given the function $y=-2-4x^2$ and the point $P(3,-38)$.

Step 2 ：To find the slope of the curve at the given point, we need to take the derivative of the function.

Step 3 ：The derivative of $y=-2-4x^2$ is $y^{\prime }=-8x$.

Step 4 ：Evaluating this derivative at the point $x=3$ gives us the slope of the curve at that point, which is $-8\ast 3=-24$.

Step 5 ：So, the slope of the curve at the point $P(3,-38)$ is $-24$.

Step 6 ：To find the equation of the tangent line at $P$, we can use the point-slope form of a line, which is $y-y_1=m(x-x_1)$, where $m$ is the slope and $(x_1,y_1)$ is the point.

Step 7 ：Substituting $m=-24$, $x_1=3$, and $y_1=-38$ into the point-slope form gives us the equation of the tangent line: $y-(-38)=-24(x-3)$.

Step 8 ：Simplifying this equation gives us $y=-24x+72$.

Step 9 ：Final Answer: The slope of the curve at the point $P(3,-38)$ is $\boxed{{\displaystyle -24}}$. The equation of the tangent line at $P$ is $\boxed{{\displaystyle y=-24x+72}}$.