Need help finding the directional derivative

Let f(x,y)=ln(x^2+y^2+1)+e^(2xy).
Find the gradient of f at the point (0,-2)
Find the directional derivative of f at the point (0,-2) in the direction of v=5i-12j
Find the maximum value of the directional derivative at the point (0,-2)

f(x,y)=ln(x²+y²+1) +e^(2xy)

∂f/∂x = 2x/x²+y²+1 +2ye^(2xy)
∂f/∂y = 2y/x²+y²+1 +2xe^(2xy)

at the point (0,-2):

∂f/∂x = -4
∂f/∂y = -4/5

so the gradient is (-4,-4/5)
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no we have to find the directional derivative in the direction of the vector v=5i-12j

a=5 , b=-12

√a²+b² = > √5²+(-12)² =>√169 =>13

u=<5/13,-12/13> =>a=5/13 , b=-12/13

Duf(x,y) = a*∂f/∂x +b*∂f/∂y

remember that the values of the partial derivatives we have already found in the first question.

Duf(x,y) = (5/13)*(-4) +(-12/13)*(-4/5) = -20/13 +48/65 =>-4/5

final answer is -4/5
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∂f/∂x = -4
∂f/∂y = -4/5

for the maximum value: √a²+b²

√(-4)²+(-4/5)² =>4√26 / 5

to find a gradient vector, you put the partial derivative for x on the left component and the partial derivative for y on the right component.

in this case, your gradient calculation will involve the chain rule. remember that the derivative of ln(u) is u'/u and (e^u)' = u'e^u

then once you find the vector, just plug in (0,-2) to answer the first part

for the second part, take the dot product of the vector you got from the first part and the unit vector of you directional vector (find the unit vector of 5i-12j first)

for the third part the gradient vector always the path of greatest rate of change, so just find the magnitude of the vector you found in the first part.