Answer
$y=1$
Work Step by Step
$y^{3}+xy^{2}+1= x+2y^{2}$
Find the y value: $y^{3}+2y^{2}+1= 2+2y^{2}$
$y=1$
Calculate by implicit differentiation
$3y^{2}\frac{dy}{dx}+2xy\frac{dy}{dx}+y^{2}=1+4y\frac{dy}{dx}$
$(3y^{2}+2xy-4y)\frac{dy}{dx}=1-y^{2}$
$\frac{dy}{dx}=\frac{1-y^{2}}{3y^{2}+2xy-4y}$
$m=0$
The equation of the tangent line is then found by using the point-slope form of the equation of a line.
$y-y_{1}=m(x-x_{1})$
$y-1=0(x-2)$
$y=1$
