5. g(x,y) = f(x,y) + 2y

Letd denote the partial derivative

Since you are on a level curve dg =dg/dx dx +dg/dy dy = 0

You obtain then:

dy/dx = -dg/dx/[dg/dy] at the point (0,1)

Now you have essentially a calculus 1 problem of finding the tangent

line to an implicitly defined curve at the point (0,1).

y = mx +1 where m = dy/dx

6. Like any directional derivative use grad(g(0,1))*

You already havedg/dy anddg/dx from part 5.