Differentiation to Find slope if tangent line Implicitly

Question

I have the equation $x^3+y^3-4xy=8$. I need to find the equation for the tangent line at $(2,0)$. When I derived the equation I came up with $$y'=\frac{3x^2-4x}{-3y^2-4y}$$

Obviously, if you plug in $x=2$ and $y=0$, your answer will be undefined. I think I differentiated improperly, can someone point out my mistake and enlighten me to the steps of proper differentiation for this equation?

Answer 1

When the derivative is taken properly you should get $y'=(4y-3x^2)/(3y^2-4x)$. Expanded it would be $3x^2+3y^2*y'-4xy'+4y=0$. By isolating the $y'$ and you get $y'(3y^2-4x)=4y-3x^2$. Divide both sides by $3y^2-4x$ and you get the proper derivative.

Answer 2

Use implicit differentiation method to find $y'$ $$x^3+y^3-4xy=8$$ $$3x^2+3y^2y'-4(y+xy')=0$$ $$3x^2+3y^2y'-4y-4xy'=0$$ $$y'(3y^2-4x)=4y-3x^2$$ $$y'=\frac{4y-3x^2}{3y^2-4x}$$ Plug in $x=2$ and $y=0$. $$y'=\frac{-12}{-8}=\frac{3}{2}=k$$ Let's find $l$. $$y=kx+l$$ $$0=3+l \to l=-3$$ Tangent line is: $y=\frac{3}{2}x-3$