I don't know E [ been too long since I took differential equations ] but I think I might be able to help you with the rest
(D) I believe that the function is a hyperbola with an equation of
y^2 - x^2 = 3
Using implicit differentiation we have that
2y y' - 2x = 0
2y y' = 2x
y' = 2x / 2y
y' = x / y
(A ) at (1,2) the slope is (x / y) = (1/2)
So the equation of the tangent line at this point is
y = (1/2) (x - 1) + 2
y = (1/2)x + 3/2
(B) In terms of x
f(x) = ±√[3 + x^2]
(C) the domain of f(x) = (-inf, inf)
Here's the graph here : https://www.desmos.com/calculator/fdxb2d50kv