Calculus 2, Is this a differential equation problem?

[∫ f(x) dx] [∫ 1/f(x) dx] = -1 solve for the second integral ∫ 1/f(x) dx let z = 1/f(x) dz/dx = -f(x)/f'(x) dx = -f'(x)/f(x) dz dx = -z f'(x) dz ∫ 1/f(x) dx = - ∫ z² f'(x) dz [∫ f(x) dx] [∫ 1/f(x) dx] = -1 [∫ f(x) dx] [- ∫ z² f'(x) dz] = -1 [∫ f(x) dx] [∫ z² f'(x) dz] = 1 ∫ z² f'(x) dz .......... integrate by parts: = uv - ∫ v du = z²f(x) - ∫ f(x) (2z dz) = 1/f(x) - 2 ∫ z f(x) dz = 1/f(x) - 2 ∫ z (1/z) dz = 1/f(x) - 2 ∫ dz = 1/f(x) - 2z + C = 1/f(x) - 2f(x) + C ∫ f(x) dx = ∫ (1/z) (-z)f'(x) dz = ∫ f'(x) dz = z f'(x) = f'(x)/f(x) [∫ f(x) dx] [∫ z² f'(x) dz] = 1 [f'(x)/f(x)] [ 1/f(x) - 2f(x) ] = 1 f'(x) ( 1 - 2[f(x)]² ) = f(x) let y = f(x) ==> y' = f'(x) y' (1 - 2y²) = y y' = y/(1 - 2y²) dy/dx = y/(1 - 2y²) (1 - 2y²)/y dy = dx (1/y - 2y) dy = dx ∫ (1/y - 2y) dy = ∫ dx ∫ 1/y dy - 2 ∫ y dy = ∫ dx lny - y² = x + C lny = y² + x + C y = e^(y² + x + C) y = A e^(x + y²) ......... where A is a constant I don't know how to solve for y, I think it got something to do with W-Lambert function or stuff like that.... hope someone can continue from here (assuming that I didn't mess up my working up there)

By inspection, the required function is : ƒ(x) = ℯˣ ......... Ans.  i.e., ℯ raised to x. _____________________________________ I am working on a proof. _____________________________________ EDIT : I just came back and saw that Hy below has given . . . . . an excellent answer to the question with a nice . . . . . proof. . . . . . The only thing for me to do now is . . . . . to give Hy a High Thumbs-Up ! ......................................…