Differential Equations Question

Hi!
I have become a bit stuck on this question the C3C4 Elmwood textbook.

Given x and y are positive, dy/dx +y/x^2 =0 and y= e when x=1, show that the solution may be written in the form y=e^(x-1)

My attempt:
dy/dx = -y/x^2
integral of 1/y with respect to y = integral of -1/x^2 with respect to x
lny = 1/3x^(-3) +c
y= Ae^(1/3x^(-3))
When y=e and x=1
e = Ae^(1/3)
A=e / e^(1/3)
A=e^(1-1/3)
A=e^(2/3)

Therefore
y=e^(1/3)*e^(1/3x^(-3))
y=e^(2/9x^3)
Which is not what I am supposed to get :/

Any suggestions to where I have gone wrong?

please post a photo of the workings

Of course

Of course

When integrating you need to raise the power by one and then divide, careful with the negative power - see your third and fourth lines of working.