2. (14 pts) Consider the function $f(x, y)=x^{3} y$
a. Find the directional derivative of $f(x, y)$ at the point $(2,1)$ in the direction from $(2,1)$ to $(3,5)$.
b. Find the direction in which $f(x, y)$ changes most rapidly at the point $(1,-1)$.
c. Find the maximum rate of change of $f(x, y)$ at the point $(1,-1)$.

Step 1 ：First, we need to find the gradient of the function \(f(x, y)=x^{3} y\) at the point \((2,1)\). The gradient of a function is a vector that points in the direction of the greatest rate of change of the function at a given point, and its magnitude is the rate of change in that direction. The gradient is given by the vector of partial derivatives of the function with respect to each variable. In this case, we need to find the partial derivatives of \(f(x, y)\) with respect to \(x\) and \(y\) at the point \((2,1)\).

Step 2 ：The partial derivative of \(f(x, y)\) with respect to \(x\) is \(3x^{2}y\), and the partial derivative with respect to \(y\) is \(x^{3}\). Evaluating these at the point \((2,1)\), we get \(f_x = 12\) and \(f_y = 8\). So, the gradient of \(f\) at \((2,1)\) is the vector \([12, 8]\).

Step 3 ：Next, we need to find the unit vector in the direction from \((2,1)\) to \((3,5)\). This is given by the difference between the coordinates of the two points, divided by the magnitude of this difference. The difference between the coordinates is \([1, 4]\), and the magnitude of this difference is \(\sqrt{17}\). So, the unit vector in the direction from \((2,1)\) to \((3,5)\) is \([\frac{\sqrt{17}}{17}, \frac{4\sqrt{17}}{17}]\).

Step 4 ：Finally, the directional derivative of the function in the direction of the unit vector is given by the dot product of the gradient and the unit vector. The dot product of \([12, 8]\) and \([\frac{\sqrt{17}}{17}, \frac{4\sqrt{17}}{17}]\) is \(\frac{44\sqrt{17}}{17}\).

Step 5 ：Final Answer: The directional derivative of \(f(x, y)=x^{3} y\) at the point \((2,1)\) in the direction from \((2,1)\) to \((3,5)\) is \(\boxed{\frac{44\sqrt{17}}{17}}\).