Question

What do I do after I find the implicit differentiation?

Answer 1

Uhm 5+8*6; uhm yeah thats all ik

Answer 2

for the implicit integration I got:

$1+3y^2y'-y'=0\\y'=-\frac{1}{3y^2-1}$

the vertical tangent would at points at which y' goes to infinity.

that happen when the right-hand side has singularities, that is, the denominator is 0:

$3y^2-1 =0\\y^2=1/3\\|y|=\sqrt{1/3}\\y=\pm\sqrt{1/3}$

so the two points y above have a vertical tangent.