5. g(x,y) = f(x,y) + 2y
Let d 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 have dg/dy and dg/dx from part 5.