f(x,y)=e^(x-√y)
(a) Square root requires y≥0. Real domain is x-√y for y≥0.
{ ∀x∈(-∞,∞),∀y∈[0,∞) }.
(b) Range: { f∈(0,∞) }. f cannot be negative or zero, but is otherwise unbounded.
(c) Real x is unbounded, but real y is limited to positive numbers, since only square roots of positive numbers (including zero) exist. On the grid the domain would be the whole area on and below the x-axis.
(d) z=f(x,y)=1, so x-√y=0, x=√y (equation) is the upper half of sideways parabola x²=y. Vertex is at the origin, passing through (1,1), (2,4), (3,9) etc.
z=f(x,y)=e, so x-√y=1 (equation is x=1+√y) is the upper half of sideways parabola y=(x-1)², vertex (1,0), x intercept = y intercept =1, passing through (2,1), (3,4), (4,9) etc.
z=f(x,y)=1/e=e⁻¹, so x-√y=-1 (equation is x=√y-1), upper half of sideways parabola y=(x+1)², vertex (-1,0), x intercept =-1, y intercept =1, passing through (1,4), (2,9), (3,16) etc.
(e) Should this reference be to part d rather than part b? Only part d includes level curves to be labelled. Use the parabola descriptions in part d to place on the grid labelling them appropriately with their equations and/or z values. The points the semi-parabolas pass through will help you to draw them (only the top half of the parabola, remember!).