If ey + xy = e, the ordered pair (dy/dx, d²y/dx²) at x = 0 is equal to?

Correct option is A (-1/e, 1/e²)
we have ey + xy = e .. (1)
from equation (1) x = 0 ⇒ y = 1
Differentiate w.r.t. x
ey(dy/dx) + x(dy/dx) + y = 0 .. (2)
dy/dx(x + ey) = -y,
dy/dx = -y/(x + ey)
Differentiate equation (2) w.r.t. x
ey * (d²y/dx²) + (dy/dx) * ey * (dy/dx) + x * (d²y/dx²) + (dy/dx) + (dy/dx) = 0
(x + ey)(d²y/dx²) + (dy/dx)² * ey + 2(dy/dx) = 0
Putting x = 0, y = 1 and using equation (3)
e(d²y/dx²) + (1/e)²e + 2(-1/e) = 0
∴ d²y/dx² = 1/e².
Hence, Option (A) is correct.