You need a point at which to find the tangent line.
Differentiate:
2x/a2+(2y/b2)dy/dx=0.
The slope of the tangent line is dy/dx=(2x/a2)/(2y/b2)=b2x/(a2y).
To find the tangent line at a point (p,q) we first find the gradient=b2p/(a2q).
The gradient must pass through (p,q) so:
y-q=(b2p/(a2q))(x-p), y=b2px/(a2q)-b2p2/(a2q)+q,
y=(b2px-b2p2+a2q2)/(a2q).
a, b, p, q are all constants so we end up with the line y=mx+c where:
m=b2p/(a2q), c=(a2q2-b2p2)/(a2q) (which is the y-intercept).
There are 4 special tangent lines at (a,0), (-a,0), (0,b), (0,-b) which are vertical and horizontal lines:
x=a, x=-a, y=b, y=-b.