just plug and chug:
4(x^2 + y^2)(2x + 2yy') = 9(2x - 2yy')
Now, the tangent is horizontal when y' = 0
4(x^2 + y^2)(2x) = 9(2x)
8x^3 + 8xy^2 = 18x
2x(4x^2 + 4y^2 - 9) = 0
So, either x=0
substitute back into original equation:
2y^4 = -9y^2
4x^2 + 4y^2 = 9
x^2 = (9 - 4y^2)/4
Substitute that back into the original equation
2((9 - 4y^2)/4 + y^2)^2 = 9((9 - 4y^2)/4 - y^2)
Expand the binomial and solve for y^2
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