Answer
$y=\dfrac{\sqrt{3}x+16\sqrt{3}}{6}.$
Work Step by Step
$\dfrac{d}{dx}(x^2y^2)-\dfrac{d}{dx}(9x^2)-\dfrac{d}{dx}(4y^2)=0\rightarrow$
$(2y^2x+\dfrac{dy}{dx}(2x^2y))-18x-\dfrac{dy}{dx}(8y)=0\rightarrow$
$y^2x-9x=\dfrac{dy}{dx}(4y-x^2y)\rightarrow$
$\dfrac{dy}{dx}=\dfrac{y^2x-9x}{4y-x^2y}.$
At $(-4, 2\sqrt{3})\rightarrow\dfrac{dy}{dx}=\dfrac{(2\sqrt{3})^2(-4)-9(-4)}{4(2\sqrt{3})-(-4)^2(2\sqrt{3})}=\dfrac{\sqrt{3}}{6}.$
Equation of tangent:
$(y-y_0)=m(x-x_0)$ at point $(x_0, y_0)$ and slope $m$.
$(y-2\sqrt{3})=\dfrac{\sqrt{3}}{6}(x+4)\rightarrow y=\dfrac{\sqrt{3}x+16\sqrt{3}}{6}.$
