#### Answer

$y=\frac{-4}{9}x+\frac{35}{9}$

#### Work Step by Step

$y^{3}+2x^{2}y-8y= x^{3}+19$
Find the y value: $y^{3}+4y-8y= 8+19$
$y=3$
Calculate by implicit differentiation
$3y^{2}\frac{dy}{dx}+2x^{2}\frac{dy}{dx}+4xy -8\frac{dy}{dx}=3x^{2}$
$(3y^{2}+2x^{2}-8)\frac{dy}{dx}=3x^{2}-4xy$
$\frac{dy}{dx}=\frac{3x^{2}-4xy}{3y^{2}+2x^{2}-8}$
$m=\frac{-4}{9}$
The equation of the tangent line is then found by using the point-slope form of the equation of a line.
$y-y_{1}=m(x-x_{1})$
$y-3=\frac{-4}{9}(x-2)$
$y=\frac{-4}{9}x+\frac{35}{9}$