find the equation of the tangent to y=x-(a^3/x^2) at the point (a,0)

how to i work this out??can someone please show me how to do it step by step?thanks!!!

Take the dervative of y

y= x- a^3 x^-2
y'=1+2a^3/x^3

so the equation for the tangent is

y= mx + b where m is y' above evaluated at (a,0)

solve for b at that tangent point.

y= x- a^3 x^-2
y'=1+2a^3/x^3
how do i work out the dervative of y?

I just did it for your, y' is the derivative.

Sorry! i mean how do you know the derivative of y= x- a^3 x^-2 is y'=1+2a^3/x^3?

i know how to change y=x-(a^3/x^2) to y= x- a^3 x^-2 but then i got lost.