Find the equation of the tangent line at the given point.?

(x^2+y^2)^2 = (25/2) x y^2

Take the derivative with respect to x
2 (x^2+y^2) (2x+2y dy/dx) = (25/2) y^2 + (25/2) x (2y) dy/dx
2 (x^2+y^2) (2x+2y dy/dx) = (25/2) y^2 + 25 x y dy/dx
2( 2x^3+2x^2y dy/dx + 2xy^2 + 2y^3 dy/dx) = (25/2) y^2 + 25 x y dy/dx
( 2x^3+2x^2y dy/dx + 2xy^2 + 2y^3 dy/dx) = (25/4) y^2 + (25/2) x y dy/dx

dy/dx ( 2x^2y + 2y^3 -(25/2) xy ) = (25/4)y^2 -2x^3-2xy^2
dy/dx = ( (25/4)y^2 -2x^3-2xy^2 ) / ( 2x^2y + 2y^3 -(25/2) xy )

substitute x=2, y=1
dy/dx = 11/12

Equation of tangent at (2,1):
y-1 = (11/12) (x-2)

y = (11/12) x - (22/12-1)
y = (11/12) x - (10/12)
y = (11/12) x - 5/6

diff both sides wrt x
2(x^2+y^2)(2x+2y dy/dx)=12.5(2xy dy/dx + y^2 )
now sub 2,1 to x,y
2(4+1)(4+2 dy/dx)=12.5(4 dy/dx +1 )
10(4+2 dy/dx)=50 dy/dx +12.5
40+20 dy/dx =50 dy/dx +12.5
dy/dx = 275/300
slope=275/300