If xy + e^y = e, find the values of y'' at the point where x = 0

Answer 1

(0,1) is on the graph
now for the derivatives. using implicit differentiation,
y + xy' + e^y y' = 0
y' = y/(x + e^y)
y" = (y'(x + e^y) - y(1 + e^y y'))/(x + e^y)^2
At (0,1), y' = 1/e
y" = ((1/e)(0+e) - 1(1 + e * 1/e))/(0 + e)^2 = -1/e^2

Answer 2

The value for y' is -y/(x+e^y)
And therefore the value of y'' is 1/e^2
Please correct the error