z=f(x,y)=x²-3xy+4y².
(a) ∂z/∂x=2x-3y, ∂z/∂y=-3x+8y.
At (-3,4), ∂z/∂x=-6-12=-18 and ∂z/∂y=9+32=41.
Derivative of z=<-18,41> at <-3,4>.
(b) At (2,1), ∂z/∂x=4-3=1, ∂z/∂y=-6+8=2,
∂z/∂u=(∂z/∂x)(∂x/∂u)=∂x/∂u,
∂z/∂u=(∂z/∂y)(∂y/∂u)=2∂y/∂u.
If Dᵤf(2,1)=3, then ∂x/∂u+2∂y/∂u=3⇒
∂x/∂u=1 and ∂y/∂u=1, u=x+p(y)=y+q(x)⇒u=x+y.
Therefore u(x,y)=x+y. Direction vector <1,1>?