You got this 4(x^2 + y^2)(2x - 2yy') = 25(2x - 2yy') I think the 2nd factor on the left-hand side should be (2x+2y y'). So it should be 4(x^2 + y^2)(2x + 2yy') = 25(2x - 2yy') Now solve for y'
you get
y′=−4x3−4xy2+25x25y+4y(x2+y2)=0 It looks like we can solve the numerator for the x values that make it zero.