Answer
$y=-\frac{9}{13}x+\frac{40}{13}$
Work Step by Step
$2(x^2+y^2)^2=25(x^2-y^2)\\
4(x^2+y^2)(2x+2yy')=25(2x-2yy')\\
4(x^2+y^2)(x+yy')=25(x-yy')\\
4x^3+4x^2yy'+4y^2x+4y^3y'=25x-25yy'\\
4x^2yy'+4y^3y'+25yy'=25x-4x^3-4y^2x\\
y'(4x^2y+4y^3+25y)=25x-4x^3-4y^2x\\
y'=\frac{25x-4x^3-4y^2x}{4x^2y+4y^3+25y}$
Plug in $(3,1)$ into $y′$ to find the gradient
$y'=\frac{25(3)-4(3)^3-4(1)^2(3)}{4(3)^2(1)+4(1)^3+25(1)}\\
y'=-\frac{9}{13}$
Equation of the tangent line at $(3,1)$
$y-1=-\frac{9}{13}(x-3)\\
y=-\frac{9}{13}x+\frac{40}{13}$
