Answer
$y=\dfrac{30-2x}{11}$
Work Step by Step
$\dfrac{d}{dx}(3(x^2+y^2)^2)=\dfrac{d}{dx}(100x^2)-\dfrac{d}{dx}(100y^2)\rightarrow$
Using the Chain Rule with $u=x^2+y^2\rightarrow\dfrac{du}{dx}=2x+\dfrac{dy}{dx}(2y)\rightarrow$
$\dfrac{d}{dx}(3(x^2+y^2)^2)=3(2x+\dfrac{dy}{dx}(2y)(2)(y^2+x^2)$
$=12(x^2+y^2)(x+\dfrac{dy}{dx}(y))$
$12(x^2+y^2)(x+\dfrac{dy}{dx}(y))=200x-\dfrac{dy}{dx}(200y)\rightarrow$
$\dfrac{dy}{dx}=\dfrac{3x^3+3y^2x-50x}{-50y-3y^3-3x^2y}$
At $(4, 2)\rightarrow\dfrac{dy}{dx}=\dfrac{3(4^3)+3(2^2)(4)-50(4)}{-50(2)-3(2^3)-3(4^2)(2)}=-\dfrac{2}{11}$
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-2)=-\frac{2}{11}(x-4)\rightarrow y=\dfrac{30-2x}{11}.$
