[tex]\text{Given that,}\\\\2(x^2+y^2)^2 = 25 (x^2-y^2)\\\\\implies 2\dfrac{d}{dx} (x^2 +y^2)^2 = 25 \dfrac{d}{dx} (x^2 -y^2)\\\\\implies 2 \cdot 2(x^2 +y^2) \dfrac{d}{dx}(x^2 +y^2) = 25\left[2x - 2y \dfrac{dy}{dx} \right]\\\\\implies (4x^2 +4y^2)\left(2x + 2y\dfrac{dy}{dx}\right) = 50x - 50y\dfrac{dy}{dx}\\\\\\\implies 8x^3+8x^2y \dfrac{dy}{dx} + 8xy^2 +8y^3\dfrac{dy}{dx} = 50x -50y \dfrac{dy}{dx}\\\\\\\implies 8x^2y \dfrac{dy}{dx} + 8y^3\dfrac{dy}{dx}+50y \dfrac{dy}{dx} = 50x -8x^3 -8xy^2\\\\[/tex]
[tex]\implies (8x^2y +8y^3 + 50y)\dfrac{dy}{dx} =50x -8x^3 -8xy^2\\\\\implies \dfrac{dy}{dx} = \dfrac{50x -8x^3 -8xy^2}{8x^2y+8y^3+50y}\\\\\implies \dfrac{dy}{dx}= \dfrac{25x-4x^3-4xy^2}{4x^2y+4y^3+25y}\\\\\\\implies \dfrac{dy}{dx} = \dfrac{x(25-4x^2-4y^2)}{y(4x^2 +4y^2 +25)}\\\\\\\text{Slope at}~~ (-3,-1)\\\\m=\dfrac{dy}{dx} = \dfrac{-3[25-4(-3)^2-4(-1)^2]}{(-1)[4(-3)^2 + 4(-1)^2+25]} = -\dfrac{45}{65} = -\dfrac 9{13}[/tex]
[tex]\text{Equation with given points,}\\\\y+1 = -\dfrac9{13}(x+3)\\\\\implies y = -\dfrac 9{13} x - \dfrac {27}{13} -1\\\\\implies y= -\dfrac 9{13} x - \dfrac{40}{13}\\\\\text{Hence the equation of this tangent line is now in a form of y= mx+b,}\\\\\text{where}~ m = -\dfrac 9{13}~ \text{and}~ b = -\dfrac{40}3[/tex]
