Find any local maximum or minimum points of the given function.
\[
y=-x^{3}+4 x^{2}+3 x+1
\]

Select the correct choice below and, if necessary, fill in the answer box to complete your choice.
A. The local maximum point(s) is/are $\square$.
(Simplify your answer. Type an ordered pair, using integers or fractions. Use a comma to separat as needed.)
B. There is no maximum.

Step 1 ：Given the function \(y=-x^{3}+4 x^{2}+3 x+1\), we need to find any local maximum or minimum points.

Step 2 ：To do this, we first find the derivative of the function. The derivative of a function gives us the slope of the function at any given point. The local maximum or minimum points occur where the derivative is zero (since the slope of the function is zero at these points).

Step 3 ：The derivative of the function \(y=-x^{3}+4 x^{2}+3 x+1\) is \(y'=-3x^{2}+8 x+3\).

Step 4 ：We set the derivative equal to zero and solve for x to find the critical points: \(x=-1/3, 3\).

Step 5 ：We substitute these x-values into the original function to get the corresponding y-values: \(y=13/27, 19\).

Step 6 ：We then determine whether these points are local maximums or minimums by taking the second derivative of the function and evaluating it at the critical points. The second derivative of the function is \(y''=8 - 6x\).

Step 7 ：By substitifying the critical points into the second derivative, we find that \((-1/3, 13/27)\) is a local minimum and \((3, 19)\) is a local maximum.

Step 8 ：Final Answer: The local maximum point is \(\boxed{(3, 19)}\).