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 $(3,1)$
$y-1=-\frac{9}{13}(x-3)\\ y=-\frac{9}{13}x+\frac{40}{13}$