Local linearization and directional derivatives

Here's another question:

""Let f(x,y)=2(x^2)y-(y^3)+x-4.

a) Find the local linearization of f at (3,1)

b) Compute the directional derivative of f at (3,1) in the direction towards the point (-2,3). At (3,1), is the function f increasing or decreasing in the direction towards the point (-2,3)? Give a reason for your answer.

c) What direction, given as a unit vector, gives the maximum instantaneous rate of change in the function at (3,1)? In what direction, given as a unit vector, is the instantaneous rate of change at (3,1) equal to 0?"

My work:

a) Local linearization: f(x,y) + fx(x,y)(x-a) + fy(x,y)(y-b)

fx(x,y)= 4xy+1 ; fx(3,1) = 13

fy(x,y) = 2x^2-3y^2 ; fy(3,1) = 15

So, the local linearization is: f(x,y) = 16 + 13(x-3) + 15(y-1)

b) u = (-2,3) ; u = 1/sqrt(2^2+3^2) = -2/sqrt(13), 3/sqrt(13) or -1/sqrt(13)

Directional derivative: (-2/sqrt(13))(13) + (3/sqrt(13))(15) = -3.05

So, the function in the direction of (3,1) is decreasing, since the rate of change is negative.

c) gradf(3,1) = 13i+15j = (13, 15) [From what I understand, the greatest increase of instantaneous change is in the direction of the gradient, so I'm assuming this is what's being looked for in terms of an answer.]

For the rate to be zero: gradf(3,1) (dot product) u = 0, so 13(x-3)+15(y-1)=0.

13x+15y=54, so anywhere on this plane is where the instantaneous rate is 0.

Did I make any mistakes?