I'm not sure how to help, because I don't know what method you used to get the tangent plane equation. What I would do would be to start with the point

on the torus, noticing that the given point

corresponds to taking

and

. Then I took the partial

and

partial derivatives of

, getting

and

as the tangent vectors in the

and

directions respectively. Next, I formed the cross product of the two tangent vectors to get a normal vector to the tangent plane. After cancelling a factor of

, I found that the normal vector was

. Plugging in the values

and

gives the normal vector

. So the equation of the tangent plane at that point is

const. Finally, substituting in the values

, I got the value

for the constant. So my equation for the tangent plane is

But don't take that on trust, because I could well have made mistakes in the calculations.