Show by implicit differentiation that the tangent to the ellipse:

x^(2) / a^(2) + y^(2) / b^(2) = 1

***
at the point (x0, y0) is:

[ (x0x) / a^(2) ] + [ y0y / b^(2) ]

*** . Please do note the correction "=1" done to the above
I got my gradient to be y' = -b^(2)x0 / a^(2)y0 but my tangent line was:

y = [ -b^(2)x0 / a^(2)y0 ] + [b^(2)x0^(2) / a^(2)y0^(2)]

*** We agree: and at the point we get what you wrote, but what you say is the tangent line's eq. cannot be correct since, for example, that has no term in it...! The eq. of the tangent line to the ellipse at this point is in fact: , and now just divide both sides by __and remember__ that is a point on the ellipse...and you'll get what you were asked!(Wink) Tonio
Im not sure what the problem may be as I simply subbed in the gradient and (x0,y0) points into the y = mx + c equation